Plane Geometry Proofs | Sec 4 A Maths Study Guide

Secondary 4 Additional Mathematics · Study guide

Turn given angle and length facts into complete geometric proofs with named reasons and correct triangle correspondence.

Concept → worked examples → practice24 min read · practice linked below

A proof is a chain of justified statements

This Secondary 4 guide develops proofs using angle facts, congruent and similar triangles, midpoints, parallel intercepts and circle theorems. Mark only what is given or already proved. A diagram suggests a route; it does not establish equality, parallelism or tangency.

Work backwards briefly from the target: a length equality may follow from congruence, a product from similarity, parallel lines from equal corresponding or alternate angles, and a cyclic quadrilateral from a circle-theorem converse. Then present the proof forwards, with each statement followed by its reason.

Choose a theorem with the correct conditions
Useful fact	What must be established
Corresponding or alternate angles are equal	The lines are parallel. Conversely, equality of the appropriate angles proves parallelism.
Perpendicular bisector	A point on it is equidistant from the endpoints; a point equidistant from distinct endpoints lies on it.
Angle bisector	It divides an angle into two equal angles. Do not infer equal opposite lengths without another theorem.
Congruent triangles	Use SSS, SAS, ASA/AAS, or RHS for right triangles. SSA alone is not a congruence test.
Similar triangles	Use AA, proportional SSS, or proportional SAS with the included angle equal. Match corresponding vertices in the written order.
Special quadrilaterals	Use the stated or proved type: a parallelogram has opposite sides parallel and equal; a rectangle also has right angles; a rhombus also has equal sides.

Example 1: use congruence to prove bisection

The diagonals of parallelogram $ABCD$ meet at M. Prove that M bisects AC.

Compare triangles $ABM$ and $CDM$ :

$AB=CD$ , because opposite sides of a parallelogram are equal.
$\angle BAM=\angle DCM$ , alternate angles with $AB\parallel CD$ and transversal AC.
$\angle ABM=\angle CDM$ , alternate angles with $AB\parallel CD$ and transversal BD.

Therefore $\triangle ABM\cong\triangle CDM$ by ASA. Corresponding sides give $AM=CM$ . Since M lies on AC, M is its midpoint. We used parallel sides and congruence to establish the required bisection.

Similar triangles turn angle facts into length ratios

Once two angles match, the third also matches because each triangle's angles sum to 180°. Write the triangles in corresponding order before taking ratios. If $\triangle ABC\sim\triangle PQR$ , then $AB/PQ=BC/QR=AC/PR$ . A common scale factor applies to lengths; areas use its square.

Example 2: prove a product using an altitude

In right triangle ABC, $\angle BAC=90^\circ$ . D lies on BC and $AD\perp BC$ . Prove $AD^2=BD\cdot DC$ .

Both $\angle ADB$ and $\angle CDA$ are 90°. Also $\angle BAD=\angle ACD$ : both are complementary to $\angle ABD=\angle ABC$ , using the angle sum in triangles ABD and ABC. Hence

\triangle ABD\sim\triangle CAD\quad\text{(AA)}.

The correspondence is $A\leftrightarrow C$ , $B\leftrightarrow A$ , $D\leftrightarrow D$ . Therefore

\frac{AD}{CD}=\frac{BD}{AD},\qquad AD^2=BD\cdot CD.

The lengths in these nondegenerate triangles are positive, so the cross-multiplication is valid. For instance, if $BD=4$ and $DC=9$ , the result gives $AD=6$ , not −6, because AD is a length.

The illustration uses BD = 4, DC = 9 and AD = 6. The angle argument proves the relation for every such right triangle.

Midpoint and intercept theorems

The midpoint theorem says that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. Its converse says that a line through the midpoint of one side, parallel to another side, bisects the remaining side. Both arise from similar triangles with scale factor $1/2$ .

Example 3: prove and use the midpoint conclusion

In triangle ABC, M is the midpoint of AB. The line through M parallel to BC meets AC at N. Prove N is the midpoint of AC and find MN if $BC=14$ .

Because $MN\parallel BC$ , $\angle AMN=\angle ABC$ and $\angle ANM=\angle ACB$ , corresponding angles. Thus $\triangle AMN\sim\triangle ABC$ by AA.

\frac{AN}{AC}=\frac{MN}{BC}=\frac{AM}{AB}=\frac12.

So $AN=AC/2=NC$ , proving N is the midpoint. Also $MN=14/2=7$ . Conversely, if M and N are given as both midpoints, $AM/AB=AN/AC=1/2$ with the common included angle A establishes similarity by SAS, and corresponding angles then prove $MN\parallel BC$ .

More generally, if D and E are inside AB and AC and $DE\parallel BC$ , then $\triangle ADE\sim\triangle ABC$ , so $AD/AB=AE/AC=DE/BC$ . Subtracting the corresponding fractional parts also gives $AD/DB=AE/EC$ . This is the intercept relationship.

Example 4: use a converse to prove parallel lines

In triangle ABC, D lies inside AB and E inside AC. Given $AD=6$ , $DB=9$ , $AE=8$ , $EC=12$ , prove $DE\parallel BC$ . If $BC=20$ , find DE.

The whole sides are $AB=15$ , $AC=20$ . Hence $AD/AB=6/15=2/5$ and $AE/AC=8/20=2/5$ . The included angle $\angle DAE$ equals $\angle BAC$ . Therefore $\triangle ADE\sim\triangle ABC$ by SAS similarity.

It follows that $\angle ADE=\angle ABC$ , so $DE\parallel BC$ by the converse of corresponding angles. The scale factor gives $DE=(2/5)(20)=8$ . The interior-point conditions ensure the smaller triangle uses the same rays from A; those conditions should be checked before using an intercept converse.

Explore: equal side fractions make a parallel segment

Move D and E independently on AB and AC. Compare $AD/AB$ with $AE/AC$ . Then match the fractions to produce a line parallel to BC. Both fractions start at one half, illustrating the midpoint theorem.

AD as a percentage of AB:50%AE as a percentage of AC:50%

Use the sliders or arrow keys. D and E stay inside their respective sides.

For these interior points, DE is parallel to BC exactly when AD/AB = AE/AC.

AD/AB = AE/AC = 0.5. DE is parallel to BC and DE/BC = 0.5. D and E are midpoints.

The moving diagram helps you notice the relationship. The similarity argument above is what proves it for every permitted configuration.

Circle theorems and their converses

Angles at the circumference subtended by the same chord in the same segment are equal. The angle at the centre is twice the angle at the circumference standing on the same arc. An angle in a semicircle is 90°. Opposite angles of a cyclic quadrilateral sum to 180°. A tangent is perpendicular to the radius at contact, and tangent lengths from the same external point are equal.

The tangent-chord theorem says that the angle between a tangent and a chord equals the angle subtended by that chord in the alternate segment. Choose the tangent ray carefully: the two rays make supplementary angles with the chord, and the opposite segment changes accordingly. Name the chord and the angle on the circumference explicitly.

Example 5: transfer an angle from a tangent

A, B and C lie on a circle. AT is a tangent at A, and the ray AT is chosen so that $\angle TAB$ is the angle corresponding to $\angle ACB$ in the alternate segment. Given $\angle TAB=58^\circ$ and $\angle ABC=47^\circ$ , find $\angle BAC$ .

Chord AB gives $\angle ACB=\angle TAB=58^\circ$ by the tangent-chord theorem. The angle sum of triangle ABC now gives

\angle BAC=180^\circ-47^\circ-58^\circ=75^\circ.

The transfer uses chord AB, so the angle on the circle must have arms CA and CB. It does not transfer 58° to $\angle ABC$ .

Chord AB connects the tangent angle at A with the angle at C in the alternate segment.

Converse theorems reverse the direction of an implication. For four distinct vertices in a convex quadrilateral, supplementary opposite angles prove that the vertices are concyclic. Alternatively, equal angles subtended by the same segment, with their vertices on the same side of that segment, establish concyclicity. Equal angles with the wrong endpoints do not.

Example 6: prove concyclicity before using it

A convex quadrilateral ABCD has $\angle ABC=110^\circ$ , $\angle ADC=70^\circ$ , and $\angle ABD=35^\circ$ . Prove A, B, C and D are concyclic, then find $\angle ACD$ .

The opposite angles satisfy $\angle ABC+\angle ADC=110^\circ+70^\circ=180^\circ$ . By the converse of the cyclic-quadrilateral theorem, the four vertices lie on one circle.

Now $\angle ABD$ and $\angle ACD$ stand on the same chord AD in the same segment. Therefore $\angle ACD=35^\circ$ . The order matters: the circle theorem becomes available only after concyclicity has been established.

Common mistakes and useful checks

Using the conclusion as a premise. If you must prove two lines parallel, you cannot yet use alternate angles from those lines.
Writing triangles in the wrong order. Check the three vertex matches before forming side ratios.
Calling similar triangles congruent. Equal angles establish shape, not equal size.
Quoting a circle theorem before proving the points concyclic. Use an appropriate converse first.
Confusing a theorem with its converse. “They are midpoints, so the joining line is parallel” and “the line is parallel through one midpoint, so it reaches the other midpoint” start from different facts.
Assuming a diagram is to scale. Name the given facts and give a reason for every new equality.

Practise complete geometric proofs

Use the Secondary 4 proofs in plane geometry questions. Write each proof before opening its worked solution, then compare the reasons and triangle correspondence. For an algebraic approach to lines and circles, see coordinate geometry and coordinate geometry of circles.