A triangle that looks isosceles is not yet known to be isosceles. A line that appears to pass through the middle of a side is not yet a median. A convincing sketch can help you notice a possible relationship, but a proof must explain why that relationship follows from the information actually given.
This is what makes plane-geometry proof different from measuring a diagram. The diagram suggests a route. The argument must remain valid for every nondegenerate configuration satisfying the stated conditions, not only for the particular picture on the page.
This guide teaches how to move from diagram clues to complete mathematical arguments. You will separate givens from targets, select congruence or similarity appropriately, use midpoint and tangent–chord relationships, and write a chain in which every conclusion has a reason. The worked configurations are described precisely so that you can sketch and label them yourself.
Curriculum scope: proofs in plane geometry are explicitly included in the 2026 O-Level Additional Mathematics 4049 syllabus and 2027 SEC G3 Additional Mathematics K341 syllabus. The listed proof tools include familiar geometry facts, congruence, similarity, the midpoint theorem and the tangent–chord theorem. This guide is written for that G3/O-Level proof route. Foundational geometry and reasoning also occur elsewhere in school Mathematics; the subject-level distinction does not make them exclusive to A-Math.
The questions and solutions are original teaching material, not reproduced examination questions. Unless stated otherwise, triangles are nondegenerate, a point described as being on a side lies on that segment, and quadrilaterals are convex.
Find a route: proof language · choosing a theorem · worked proofs · tangent–chord reasoning · mistake clinic · practice · full solutions.
Four kinds of information must not be mixed
A given is information supplied by the question: AB = AC, D is the midpoint of BC, or AB is parallel to CD. You may use it immediately. Mark it clearly on a sketch, but do not add further equalities merely because the drawing appears symmetric.
An established fact is something you have already justified during the solution. After proving two triangles congruent, you may use their corresponding sides or angles as equal. Before proving congruence, those conclusions are not yet available.
A theorem is an accepted result whose conditions must be satisfied. For example, the midpoint theorem requires the relevant two midpoints, not just a line that looks halfway up a triangle. The tangent–chord theorem requires the correct tangent, chord and angle in the alternate segment.
The target is what you are trying to establish. It cannot be used as a premise in its own proof. Working backwards from the target is useful planning; writing the target as an assumed fact and then deriving something known is not automatically a valid forward proof.
Read the labels exactly
In ∠ABC, the middle letter B is the vertex. The angle is formed by BA and BC. Changing the order to ∠BAC changes the vertex and usually changes the angle.
AB can denote the length of segment AB in an equation such as AB = AC. When using angle notation, keep three-letter names whenever a vertex has several rays. “Angle A” can become ambiguous after extra lines are drawn.
A midpoint statement contains two kinds of information: the point lies on the segment, and it divides the segment into two equal lengths. Saying BD = DC alone is not enough to place D on BC; many points off the line can be equidistant from B and C.
Likewise, perpendicular lines give right angles only at their intersection. Parallel lines give corresponding or alternate-angle relationships when an appropriate transversal connects them. Every theorem has a configuration, not just a name.
Plan backwards, then write forwards
Suppose the target is DE parallel to BC. Ask what would establish that. Equal corresponding angles could do it. Similar triangles might provide those equal angles. Proportional sides and an included angle might establish the similarity.
This backward plan helps you search. The written proof should then begin with the known side ratios and common angle, establish similarity, obtain the corresponding angle equality and conclude parallelism. The final chain follows from givens to target.
A two-column format can help during learning: one column for the statement and one for its reason. A polished paragraph is also acceptable when the logic is clear. The format is secondary; the important point is that the reader can see why each new claim is available.
Keep calculations and reasons close together. Writing a list of equal angles followed by a distant list of theorem names makes it hard to know which reason supports which equality.
Choose the right tool, not the most familiar word
| Situation | Possible tool | What to check |
|---|---|---|
| Need equality of corresponding parts | Triangle congruence | A valid SSS, SAS, ASA, AAS or RHS condition |
| Need proportional lengths or matching angles | Triangle similarity | AA, proportional SAS or proportional SSS |
| A line joins two side midpoints | Midpoint theorem | Both midpoint conditions and the correct triangle |
| A midpoint line is parallel to a third side | Similarity or the relevant midpoint converse | Which midpoint is given and which is the target |
| A tangent meets a chord | Tangent–chord theorem | The tangent ray and the angle in the alternate segment |
| A radius reaches a tangent point | Radius–tangent perpendicularity | The radius ends at the actual point of contact |
| Need to prove two lines parallel | Converse angle relationships | A justified equal-angle pair and its transversal |
Congruence is stronger than similarity
Congruent triangles have the same shape and size. Their corresponding sides and angles are equal. Similar triangles have the same shape, but their corresponding sides may differ by a common scale factor.
Three equal corresponding angles establish similarity, not congruence. A triangle with side lengths 3, 4 and 5 is similar to one with side lengths 6, 8 and 10, but they are not congruent. The second is twice the size in every corresponding length.
For SAS congruence, the equal angle must be included between the two stated equal sides. Two sides and an unrelated angle do not provide a general congruence rule. For RHS, first establish that both triangles are right-angled, then identify the equal hypotenuse and a corresponding equal side.
Write similarity in the correct vertex order
If triangle ADE is similar to triangle ABC in that order, then A corresponds to A, D to B and E to C. Therefore AD/AB = AE/AC = DE/BC.
Writing AD/AB = AC/AE reverses only one of the scale ratios and is generally wrong. You may invert all the corresponding ratios consistently, but you cannot invert one pair while leaving the others unchanged.
Before using a ratio, write a short correspondence: A ↔ A, D ↔ B, E ↔ C. This is particularly helpful when triangles overlap or one is rotated relative to the other.
Worked proof 1: an isosceles triangle and its median
Given: triangle ABC has AB = AC. Point D is the midpoint of BC. Prove that AD is perpendicular to BC and that AD bisects ∠BAC.
Sketch: draw triangle ABC with A above the segment BC. Place D on BC and mark BD = DC. Join A to D. The drawing may look symmetric, but the proof uses the two stated equalities rather than the appearance.
| Statement | Reason |
|---|---|
| AB = AC | Given |
| BD = DC | D is the midpoint of BC |
| AD = AD | Common side |
| Triangle ABD is congruent to triangle ACD | SSS |
| ∠BDA = ∠ADC | Corresponding angles of congruent triangles |
| ∠BDA + ∠ADC = 180° | B, D and C lie on one straight line |
| Each of those angles is 90° | They are equal and sum to 180° |
| AD is perpendicular to BC | Definition of perpendicular lines |
The same congruence gives ∠BAD = ∠DAC. Therefore AD also bisects ∠BAC. One established relationship, triangle congruence, supports two different conclusions.
Notice what was not assumed: we did not call AD an altitude at the start, and we did not use a right angle to prove the triangles congruent. The right angle was the result. Using RHS before establishing perpendicularity would reverse the logic.
Worked proof 2: a midpoint and a parallel line
Given: D is the midpoint of AB in triangle ABC. A line through D parallel to BC meets AC at E. Prove that E is the midpoint of AC and that DE = BC/2.
Because DE is parallel to BC, ∠ADE = ∠ABC and ∠AED = ∠ACB by corresponding-angle relationships. Therefore triangle ADE is similar to triangle ABC by AA, with A ↔ A, D ↔ B and E ↔ C.
Since D is the midpoint, AD/AB = 1/2. Similarity then gives AE/AC = 1/2 and DE/BC = 1/2. Thus AE = AC/2. Because E lies on AC, EC = AC − AE = AC/2, so AE = EC. Therefore E is the midpoint of AC and DE = BC/2.
If AC = 18 and BC = 14, then AE = EC = 9 and DE = 7. These numerical consequences check the scale factor but are not substitutes for the general proof.
It would be circular to begin “D and E are midpoints” because only D was given as a midpoint. The role of the proof is to establish the missing midpoint property.
Worked proof 3: two midpoints produce a parallel line
Given: D and E are the midpoints of AB and AC respectively in triangle ABC. Prove DE parallel to BC and DE = BC/2 without assuming the conclusion.
The midpoint conditions give AD/AB = AE/AC = 1/2. The included angle ∠DAE equals ∠BAC because D lies on AB and E lies on AC. The two pairs of sides around that angle are proportional, so triangle ADE is similar to triangle ABC by SAS similarity.
Similarity gives ∠ADE = ∠ABC. These are corresponding angles for lines DE and BC with transversal AB, so DE is parallel to BC. Similarity also gives DE/BC = AD/AB = 1/2, hence DE = BC/2.
This argument establishes the midpoint theorem’s conclusion. In a question that allows the theorem directly, naming it after verifying both midpoints may be sufficient. In a question asking you to prove the result, simply citing the result being proved would not provide the intended argument.
The difference between worked proofs 2 and 3 is important. One begins with parallelism and proves the second midpoint. The other begins with two midpoints and proves parallelism. Similarity connects the facts, but the order of permission changes.
Worked proof 4: diagonals inside a trapezium
Given: ABCD is a convex trapezium with AB parallel to CD. Diagonals AC and BD meet at O. Show that AO/OC = BO/OD = AB/CD.
Sketch: draw AB as the shorter top parallel side and CD as the longer bottom side, keeping vertices A, B, C and D in order around the boundary. Draw both diagonals and label their intersection O. The two triangles to compare are AOB and COD.
∠AOB = ∠COD because they are vertically opposite angles. Also, ∠ABO = ∠CDO because AB is parallel to CD and BD is a transversal. Therefore triangle AOB is similar to triangle COD by AA.
The correspondence is A ↔ C, O ↔ O and B ↔ D. Hence AO/OC = BO/OD = AB/CD.
If AB = 6, CD = 9 and AO = 4, the scale relation gives 4/OC = 6/9, so OC = 6. If BO = 5, then 5/OD = 6/9 and OD = 7.5.
The ratio uses corresponding pieces of the diagonals, not their whole lengths. Once OC is known, AC = AO + OC = 10 because O lies between A and C. Do not replace OC by AC inside the similarity ratio.
Worked proof 5: why parallelogram diagonals bisect each other
Given: ABCD is a parallelogram, and AC and BD meet at O. Prove AO = OC and BO = OD.
Use the already established parallelogram property AB = CD. The parallel sides give ∠ABO = ∠CDO and ∠BAO = ∠DCO by alternate-angle relationships. Together with AB = CD, these establish triangle ABO congruent to triangle CDO by ASA.
Corresponding sides then give AO = CO and BO = DO. Thus O bisects both diagonals.
This proof does not begin by labelling O as the midpoint. O is initially just the intersection. Bisection is established afterwards. It also makes its use of the earlier opposite-side property explicit. A proof can use previously established theorems; it must not conceal the target among its assumptions.
Tangent–chord reasoning: choose the angle, not just the theorem
A tangent line at A contains two opposite rays. A chord AB makes supplementary angles with those rays. Unless the chord is perpendicular to the tangent, the two angles are different, so selecting the correct ray matters. In the special diameter case, both tangent–chord angles are 90°.
For the next worked example, use this precise configuration. Draw a circle above a horizontal tangent at its lowest point A. Put T on the tangent to the right of A. Choose B on the right half of the circle, away from the endpoints of its vertical diameter, and choose C on the major arc AB, distinct from A and B. Join AB, AC and BC. The tangent–chord theorem relates ∠TAB to ∠ACB in this configuration.
This construction identifies the relevant ray and segment rather than relying on a vague instruction to “use the tangent theorem”. When a supplied diagram uses a different orientation, inspect its actual rays and arcs before carrying an angle across.
Worked proof 6: a tangent angle followed by a triangle angle
In the configuration just described, ∠TAB = 38° and ∠ABC = 67°. Find ∠ACB and ∠BAC, giving reasons.
By the tangent–chord theorem, ∠ACB = ∠TAB = 38°. The angles of triangle ABC sum to 180°, so ∠BAC = 180° − 38° − 67° = 75°.
Now put S on the same tangent line to the left of A. Rays AS and AT are opposite, so ∠SAB + ∠TAB = 180°. Therefore ∠SAB = 142°, not 38°.
The first equality uses a circle theorem; the last uses a straight-line relationship. Naming “tangent–chord” beside every angle would hide which configuration actually justifies each step.
A useful check is that the triangle angles 38°, 67° and 75° add to 180°. This arithmetic check can expose a wrong angle transfer, but the theorem and ray choice are still what justify the result.
Worked proof 7: two tangents from one external point
Given: T is outside a circle with centre O. TA and TB are the two tangents, meeting the circle at A and B. Prove TA = TB and that OT bisects ∠ATB.
Join OA, OB and OT. A radius is perpendicular to the tangent at its point of contact, so ∠OAT = ∠OBT = 90°. OA = OB because they are radii of the same circle. OT is the common hypotenuse of right-angled triangles OAT and OBT.
Therefore triangle OAT is congruent to triangle OBT by RHS. Corresponding sides give TA = TB, and corresponding angles at T give ∠ATO = ∠OTB. Thus OT bisects ∠ATB.
For a numerical check, suppose OA = 5 and OT = 13. Pythagoras in right-angled triangle OAT gives TA2 = 132 − 52 = 144, so TA = 12. The congruence result then gives TB = 12.
If the question asks you to prove the equality of the tangents, writing “TA = TB because tangents from an external point are equal” simply names the target theorem. The RHS route explains why the result holds.
Worked proof 8: a circle fact creates an isosceles triangle
Use the same tangent–chord configuration as worked proof 6, but now suppose the given condition is ∠TAB = ∠ABC. Prove AB = AC.
The tangent–chord theorem gives ∠TAB = ∠ACB. Combining this with the given equality gives ∠ABC = ∠ACB. Therefore the sides opposite those equal angles are equal: AC = AB.
This is a short proof, but it contains three distinct actions: transfer an angle through a circle theorem, combine two equalities, and apply the converse of the isosceles-triangle angle result.
There is no reason to search for a complicated congruence argument once those equal angles are available. A strong proof selects enough machinery to establish the target and then stops.
How an added line can help without changing the givens
Joining two existing points is often a legitimate construction that makes an available relationship visible. In the two-tangent proof, joining the centre to the contact points creates right-angled triangles. In the isosceles-median proof, joining A to D creates two triangles sharing a side.
The construction itself does not grant new equalities. Drawing AD does not make it perpendicular or an angle bisector. Those facts still require proof. Likewise, drawing a line that appears parallel does not authorise parallel-angle rules unless it is explicitly constructed parallel or subsequently proved parallel.
Choose an added line because it serves a target: a common side for congruence, a radius at a tangent, or a transversal between known parallel lines. Adding many unexplained lines can make the diagram harder to read without adding evidence.
Mistake clinic: plausible statements that do not yet prove anything
“DE looks parallel to BC.” Appearance is a clue, not a reason. Establish equal corresponding or alternate angles, or verify the conditions of an appropriate theorem. A sketch can be intentionally not to scale.
“The triangles have three equal angles, so they are congruent.” They are similar. The 3–4–5 and 6–8–10 triangles provide a counterexample to congruence. To prove congruence, you need enough size information as well.
“D is a midpoint, so DE is a midline.” One midpoint alone does not establish the second endpoint or the line’s direction. In worked proof 2, parallelism was additionally given and similarity established E as the second midpoint.
“The two right angles look obvious.” State why they are right angles. Radius–tangent perpendicularity, a given perpendicular line, or an already proved angle result may supply the reason. A right-angle appearance does not qualify a triangle for RHS.
“The ratio is AD/AB = AC/AE.” Check the correspondence. For triangle ADE similar to ABC, both ratios should compare the smaller triangle with the larger: AD/AB = AE/AC. Inverting only one ratio breaks that consistency.
“Assume DE is parallel to BC; therefore the given ratios hold; so DE is parallel.” This has used the target as a premise. A valid reverse-direction argument must start with the actual given ratios and establish the appropriate angle equality or similarity.
“The tangent angle is 38° on either side.” Opposite tangent rays make supplementary angles with the chord. Identify the ray before applying the alternate-segment relation.
“The numerical example worked, so the theorem is proved.” A calculation for one triangle tests that case. A general proof must use relationships that hold for every allowed configuration. Numbers can reveal an error or illustrate a theorem, but a few successful measurements do not establish universality.
Independent practice: twelve tasks
Draw your own labelled sketch for each configuration. Mark only the givens initially. As you establish a new fact, add it with its reason. Keep the target visible so that your proof does not wander into unrelated geometry.
1. A triangle ABC is drawn symmetrically, but no equality marks or length statements are provided. May you conclude AB = AC? Explain.
2. In triangle ABC, AB = AC and ∠BAC = 40°. Find ∠ABC and ∠ACB with reasons.
3. D and E are the midpoints of AB and AC in triangle ABC. If BC = 18, state DE and its relationship to BC, naming the theorem.
4. Triangle ADE is similar to triangle ABC in that order. AD = 3, AB = 9, AE = 4 and BC = 15. Find AC and DE.
5. In convex trapezium ABCD, AB is parallel to CD and diagonals meet at O. Given AB = 4, CD = 10, AO = 6 and BO = 8, find OC and OD. Establish the similarity used.
6. Prove that the median from the apex of an isosceles triangle is perpendicular to its base. Use congruence rather than assuming a right angle.
7. D lies inside AB and E inside AC in triangle ABC. Given AD/AB = AE/AC = 2/3, prove DE parallel to BC. If BC = 12, find DE.
8. Use the tangent–chord configuration described in worked proof 6. Given ∠TAB = 42° and ∠ABC = 58°, find ∠ACB and ∠BAC. If S is on the opposite tangent ray, find ∠SAB.
9. From an external point T, tangents TA and TB touch a circle of centre O and radius 5. If OT = 13, find TA and TB and justify the equality used.
10. D and E lie inside AB and AC, and triangle ADE is similar to triangle ABC in that order. A student writes AD/AB = AC/AE. Repair the ratio and explain the mistake.
11. Give a counterexample to “Two triangles with equal corresponding angles must have equal corresponding sides”.
12. ABCD is a convex cyclic quadrilateral with ∠DAB = 112° and ∠ABC = 74°. Find the interior angles at C and D, giving reasons.
Full solutions
1–3: distinguish appearance, given information and a theorem
1. No. Symmetry in a drawing does not establish equality of lengths. You need a given equality or a justified argument that leads to it. A triangle with unequal sides can be drawn approximately symmetrically.
2. AB = AC implies ∠ABC = ∠ACB because base angles opposite equal sides are equal. Their sum is 180° − 40° = 140°, so each is 70°.
3. Both midpoint conditions are supplied, so the midpoint theorem applies. DE is parallel to BC and DE = 9. State the correct triangle; a midpoint theorem applies within a particular triangle, not to any two points labelled as midpoints elsewhere.
4–5: keep corresponding ratios consistent
4. The scale factor from ABC to ADE is AD/AB = 3/9 = 1/3. Hence AE/AC = 1/3 gives 4/AC = 1/3, so AC = 12. Also DE/BC = 1/3, giving DE = 5.
5. ∠AOB = ∠COD by vertically opposite angles, and ∠ABO = ∠CDO by the parallel lines and transversal BD. Thus triangle AOB is similar to COD. The ratio is AO/OC = BO/OD = AB/CD = 4/10. Therefore 6/OC = 4/10 gives OC = 15, and 8/OD = 4/10 gives OD = 20.
6–7: establish the relationship before using its consequence
6. Label the isosceles triangle ABC with AB = AC and the base midpoint D. In triangles ABD and ACD, AB = AC, BD = DC and AD is common. They are congruent by SSS. Therefore ∠BDA = ∠ADC. These adjacent angles sum to 180° on line BC, so each is 90°. Hence AD is perpendicular to BC.
7. The two side ratios around angle A are equal, and ∠DAE = ∠BAC. Therefore triangle ADE is similar to ABC by SAS similarity. Corresponding angles ∠ADE and ∠ABC are equal, so DE is parallel to BC. Similarity gives DE/BC = 2/3, hence DE = 8.
8–9: identify the contact points and tangent rays
8. The tangent–chord theorem gives ∠ACB = 42°. The triangle angle sum gives ∠BAC = 180° − 42° − 58° = 80°. Opposite rays AS and AT give ∠SAB = 180° − 42° = 138°.
9. OA is perpendicular to TA and OB to TB because the radii meet tangents at their contact points. In right-angled triangle OAT, TA2 = 132 − 52 = 144, so TA = 12. Triangles OAT and OBT are congruent by RHS: common hypotenuse OT and equal radii OA, OB. Therefore TB = TA = 12.
10–12: repair the reasoning, not only the answer
10. The correct ratio is AD/AB = AE/AC. Each fraction compares a side of ADE with its corresponding side of ABC. The student’s second fraction reverses that direction. An equivalent consistent statement would be AB/AD = AC/AE.
11. A 3–4–5 triangle and a 6–8–10 triangle have equal corresponding angles because their corresponding sides are proportional. Their side lengths are not equal. They are similar but not congruent.
12. Opposite interior angles of a cyclic quadrilateral sum to 180°. Thus the angle at C is 180° − 112° = 68°, and the angle at D is 180° − 74° = 106°. The cyclic condition is essential to that reasoning.
A delayed retest with new labels
Retest A: In triangle PQR, S and U are the midpoints of PQ and PR. Prove SU parallel to QR using side ratios and the common included angle. Do not rely on remembering the letters from an earlier example.
The ratios PS/PQ and PU/PR are both 1/2, while ∠SPU = ∠QPR. Therefore triangle PSU is similar to PQR by SAS similarity. Corresponding-angle equality gives SU parallel to QR. The proof is structurally the same even though every label changed.
Retest B: In a correctly identified tangent–chord configuration, the angle between the selected tangent ray at A and chord AB is given equal to ∠ABC. Show that AB = AC.
The tangent–chord theorem transfers that tangent angle to ∠ACB. Hence ∠ABC = ∠ACB, and the opposite sides AC and AB are equal. The important skill is noticing which new angle equality the circle theorem creates.
After a delay, ask whether you could name the comparison triangles, establish the correspondence and choose the theorem without copying a previous proof. That is a more useful check than recognising that the finished solution looks familiar.
How to practise proof without memorising paragraphs
Begin with a target and a small set of available facts. Ask which conclusion becomes available next. Once a route is found, remove the prompts and write the proof from the original givens. This trains the decisions that a memorised paragraph can conceal.
Use contrast pairs. Compare two midpoints with one midpoint plus parallelism. Compare AAA similarity with SSS congruence. Compare the two opposite tangent rays. These pairs show exactly which condition changes what you are allowed to conclude.
When a proof fails, identify the first unsupported statement. Everything after that line may be logically dependent on it. Correcting the final numerical answer alone does not repair the argument.
A useful correction record contains the target, the first unsupported claim, the missing condition, and a repaired chain. Then change the diagram’s orientation or the letters and retest. A proof method should survive rotation, relabelling and a different-looking sketch.
For a student who writes too much, ask whether each line advances the target. For a student who writes too little, ask what justifies each equality. The aim is neither maximal length nor minimal ink; it is an argument whose necessary links are visible.
Where coordinate geometry fits—and where it does not replace proof
Some plane-geometry results can also be established using coordinates, gradients or distances. Placing a midpoint at the origin or aligning a side with an axis may simplify an argument. The Coordinate Geometry guide owns those algebraic representations.
However, coordinates must represent the full allowed configuration. Choosing convenient numerical points that accidentally make a general triangle isosceles would prove only a special case. A coordinate proof should retain enough variables and conditions to cover the intended statement.
For this guide’s main tasks, congruence, similarity and circle theorems often provide the clearest route. Choose a method because it reveals the needed relationship, not because it is the only method you remember.
A final proof checklist
Before finishing, check that every equality comes from a given, a valid theorem or an earlier established step. Check vertex order in congruence and similarity statements. Check that tangent angles use the right rays, that midpoint statements include betweenness, and that no target was quietly assumed.
Then read only the reasons. Do they form a coherent route? “Given → proportional sides and common angle → similarity → corresponding angles → parallel lines” is a complete mechanism. “Looks equal → should be similar → therefore parallel” is not.
Geometry becomes less mysterious when the student learns to ask what information a theorem requires and what new information it returns. A diagram is no longer a puzzle to guess from; it becomes a record of relationships that can be established one justified step at a time.
Use the Geometry and Trigonometry strand guide for the wider topic structure, or return to the Additional Mathematics Hub for the next learning route. The separate Additional Mathematics Tuition page explains teaching support. For a proof difficulty, bring the student’s attempted argument, not only the completed model answer.
Source and scope review: 6 September 2026. The linked official subject-content tables establish the G3/O-Level proof scope. The configurations, worked arguments, counterexamples, practice questions and solutions are original teaching material. Numerical illustrations support understanding; the written theorem-based chains provide the general arguments.