Geometric proofs - Answer key

SSS: three pairs of corresponding sides equal. SAS: two sides and the included angle equal. AAS: two angles and a corresponding side equal. RHS: right angle, hypotenuse, and one other side equal.
SAS (two sides and the included angle between them).
$\angle PQR = \angle PRQ = \frac{180° - 70°}{2} = 55°$ .
The other alternate angle is $55°$ . The co-interior angle on the same side is $180° - 55° = 125°$ .
Exterior angle $= \angle X + \angle Y$ , so $130° = 80° + \angle Y$ , giving $\angle Y = 50°$ .
A demonstration checks specific cases (e.g. measuring angles in three triangles and finding they sum to $180°$ ). A deductive proof uses logical steps from axioms to show the result holds for all cases. The demonstration could fail for an untested case; the proof cannot.
The diagonals of a parallelogram bisect each other (proven by showing two pairs of congruent triangles using SAS with vertically opposite angles).
Yes, by RHS: right angle, hypotenuse $BC = \sqrt{36 + 64} = 10 =$ hypotenuse of $\triangle DEF$ , and $AB = DE = 6$ .

Since $AM = \frac{1}{2}BC$ and $M$ is the midpoint, $BM = MC = AM$ . So $\triangle ABM$ is isosceles ( $AM = BM$ ) and $\triangle ACM$ is isosceles ( $AM = CM$ ). Let $\angle ABM = \alpha$ and $\angle ACM = \beta$ . In $\triangle ABM$ : $\angle BAM = 180° - 2\alpha$ . In $\triangle ACM$ : $\angle CAM = 180° - 2\beta$ . Also $\alpha + \beta = \angle AMB$ or more directly: $\angle BAC = \angle BAM + \angle CAM = (180° - 2\alpha) + (180° - 2\beta) = 360° - 2(\alpha + \beta)$ . But $\alpha + \beta = \angle ABC + \angle ACB = 180° - \angle BAC$ . So $\angle BAC = 360° - 2(180° - \angle BAC) = 360° - 360° + 2\angle BAC$ , giving $\angle BAC = 2\angle BAC - 0$ … Let us redo cleanly. In $\triangle ABC$ : $\angle A + \alpha + \beta = 180°$ , so $\alpha + \beta = 180° - \angle A$ . In $\triangle ABM$ (isosceles with $AM = BM$ ): $\angle BAM = \alpha$ , so $\angle AMB = 180° - 2\alpha$ . In $\triangle ACM$ (isosceles with $AM = CM$ ): $\angle CAM = \beta$ , so $\angle AMC = 180° - 2\beta$ . But $\angle AMB + \angle AMC = 180°$ (straight line). So $(180° - 2\alpha) + (180° - 2\beta) = 180°$ , giving $360° - 2(\alpha + \beta) = 180°$ , so $\alpha + \beta = 90°$ . Therefore $\angle A = 180° - 90° = 90°$ .
In kite $ABCD$ with $AB = AD$ and $CB = CD$ , let diagonals meet at $O$ . In $\triangle ABC$ and $\triangle ADC$ : $AB = AD$ , $CB = CD$ , $AC$ common. By SSS, $\triangle ABC \cong \triangle ADC$ . So $\angle BAC = \angle DAC$ . Now in $\triangle ABO$ and $\triangle ADO$ : $AB = AD$ , $\angle BAO = \angle DAO$ , $AO$ common. By SAS, $\triangle ABO \cong \triangle ADO$ . So $BO = DO$ (diagonal bisected) and $\angle AOB = \angle AOD = 90°$ (supplementary and equal).
Let the centres be $O_1$ and $O_2$ with equal radius $r$ . Then $O_1P = O_1Q = r$ and $O_2P = O_2Q = r$ . So $O_1$ and $O_2$ are each equidistant from $P$ and $Q$ , meaning both lie on the perpendicular bisector of $PQ$ . The line $O_1O_2$ is therefore the perpendicular bisector of $PQ$ .
$BD = \sqrt{AB^2 - AD^2} = \sqrt{169 - 144} = 5$ cm. $AC = \sqrt{AD^2 + DC^2} = \sqrt{144 + 16} = \sqrt{160} = 4\sqrt{10} \approx 12.65$ cm.
In rhombus $ABCD$ , all sides equal. Diagonals $AC$ and $BD$ meet at $O$ . In $\triangle AOB$ and $\triangle COB$ : $AB = CB$ (rhombus), $OB$ common, $AO = OC$ (to prove). Instead: in $\triangle ABD$ and $\triangle CBD$ : $AB = CB$ , $AD = CD$ , $BD$ common. By SSS, $\triangle ABD \cong \triangle CBD$ . So $\angle ABD = \angle CBD$ . Now in $\triangle ABO$ and $\triangle CBO$ : $AB = CB$ , $\angle ABO = \angle CBO$ , $BO$ common. By SAS, congruent. So $AO = CO$ and $\angle AOB = \angle COB = 90°$ .
Let $AD = AE$ (since $BD = CE$ and $AB = AC$ , we have $AD = AB - BD = AC - CE = AE$ ). In $\triangle ADE$ : $AD = AE$ , so $\triangle ADE$ is isosceles with $\angle ADE = \angle AED$ . Also $\angle ADE + \angle AED + \angle A = 180°$ . In $\triangle ABC$ : $\angle ABC = \angle ACB$ (isosceles) and $\angle ABC + \angle ACB + \angle A = 180°$ . So $\angle ADE = \angle ABC$ . Since these are corresponding angles with transversal $AB$ cutting $DE$ and $BC$ , we get $DE \parallel BC$ .

Incorrect. Three equal angle pairs (AAA) guarantee similarity, not congruence. For example, an equilateral triangle with side $3$ cm and an equilateral triangle with side $6$ cm both have all angles $= 60°$ but are clearly not congruent. AAA does not fix the size of the triangle.
Let $BC$ be a diameter with centre $O$ , and $A$ a point on the circle. Then $OA = OB = OC = r$ . In $\triangle OAB$ : $OA = OB$ , so $\angle OAB = \angle OBA = \alpha$ . In $\triangle OAC$ : $OA = OC$ , so $\angle OAC = \angle OCA = \beta$ . In $\triangle ABC$ : $\angle BAC = \alpha + \beta$ and $\angle ABC + \angle BCA + \angle BAC = 180°$ , i.e. $\alpha + \beta + (\alpha + \beta) = 180°$ , so $2(\alpha + \beta) = 180°$ , giving $\angle BAC = 90°$ .
In $\triangle ABC$ and $\triangle CDA$ : $AB = CD$ (given), $AC$ common, $\angle BAC = \angle DCA$ (alternate angles since $AB \parallel CD$ ). By SAS, $\triangle ABC \cong \triangle CDA$ . Therefore $BC = DA$ and $\angle BCA = \angle DAC$ (alternate angles), so $BC \parallel AD$ . Both pairs of opposite sides are parallel, hence $ABCD$ is a parallelogram.
SSA is ambiguous because the given angle is not between the two known sides. For example: $\triangle_1$ has $a = 5$ , $b = 7$ , $\angle A = 30°$ ; $\triangle_2$ has $a = 5$ , $b = 7$ , $\angle A = 30°$ but $\angle B$ can be either acute or obtuse (since $\sin B = \frac{7\sin 30°}{5} = 0.7$ gives $B \approx 44.4°$ or $B \approx 135.6°$ ). Two different triangles satisfy the same SSA conditions.

In $\triangle ABC$ with $\angle A = 60°$ and $AB = AC = 10$ , the triangle is isosceles. By the cosine rule: $BC^2 = 100 + 100 - 2(100)\cos 60° = 200 - 100 = 100$ , so $BC = 10$ . The triangle is equilateral. The circumradius $R = \frac{a}{2\sin A} = \frac{10}{2\sin 60°} = \frac{10}{\sqrt{3}} = \frac{10\sqrt{3}}{3} \approx 5.77$ cm. So $PA = \frac{10\sqrt{3}}{3}$ cm.
Let medians from $A$ and $B$ meet at $G$ . The median from $A$ to midpoint $M$ of $BC$ : by the vector approach, $G = A + \frac{2}{3}(M - A) = \frac{A + 2M}{3} = \frac{A + B + C}{3}$ . Similarly, the median from $B$ to midpoint $N$ of $AC$ gives the same point $G = \frac{A + B + C}{3}$ . The third median from $C$ to midpoint $P$ of $AB$ also passes through $\frac{A + B + C}{3}$ . Since all three medians yield the same intersection point, they are concurrent, and $G$ divides each median in the ratio $2:1$ from vertex.
Let $\angle AOB = 2\alpha$ and $\angle AOD = 2\beta$ where $O$ is the centre (central angles subtended by arcs $AB$ and $AD$ ). Then $\angle ACB = \alpha$ and $\angle ACD = \beta$ (angle at circumference = half central angle). Opposite angle $\angle B =$ angle subtended by arc $ADC$ at circumference. Arc $ADC$ has central angle $2\beta + 2\gamma$ (where $2\gamma$ is the central angle for arc $DC$ ). The key result: $\angle B + \angle D$ corresponds to arcs that together make a full circle ( $360°$ ), so the sum of the half-angles is $180°$ .
By the cosine rule in $\triangle ABC$ : $BC^2 = AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos A$ . Similarly $B'C'^2 = A'B'^2 + A'C'^2 - 2 \cdot A'B' \cdot A'C' \cos A'$ . Since $AB = A'B'$ and $AC = A'C'$ , and $\angle A > \angle A'$ , we have $\cos A < \cos A'$ (cosine is decreasing on $[0°, 180°]$ ). Therefore $-2 \cdot AB \cdot AC \cos A > -2 \cdot A'B' \cdot A'C' \cos A'$ , so $BC^2 > B'C'^2$ , hence $BC > B'C'$ .

Tier 1

Tier 2

Tier 3

Challenge