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Q1.

Statement (short): ∆ABC and ∆DBC are two isosceles triangles on the same base BC with vertices A and D on the same side of BC. AD is extended to meet BC at P. Prove:
(i) ∆ABD ≅ ∆ACD.
(ii) ∆ABP ≅ ∆ACP.
(iii) AP bisects ∠A and ∠D.
(iv) AP is the perpendicular bisector of BC.

Solution.

(i) From the problem: $AB=AC$ and $DB=DC$. In triangles $ABD$ and $ACD$ the three sides satisfy

$$AB=AC,BD=CD,AD=AD(\text{common})\mathrm{}$$

So by SSS, $\mathrm{△}ABD\cong \mathrm{△}ACD$

(ii) From (i) the congruence gives $\mathrm{\angle }BAD=\mathrm{\angle }DAC$. Extend AD to P (so P lies on line AD). In triangles $ABP$ and $ACP$:

• $AB=AC$ (given),

• $\mathrm{\angle }BAP=\mathrm{\angle }PAC$ (since $\mathrm{\angle }BAD=\mathrm{\angle }DAC$ and P is on the same line),

• $AP=AP$ (common).
  So by SAS, $\mathrm{△}ABP\cong \mathrm{△}ACP$

(iii) From (ii) corresponding parts give $\mathrm{\angle }BAP=\mathrm{\angle }PAC$ and $\mathrm{\angle }BPA=\mathrm{\angle }APC$. Thus $AP$bisects $\mathrm{\angle }A$. Because P lies on line AD, the same symmetry around line AP applied to the configuration with vertex $D$shows $AP$ also bisects $\mathrm{\angle }D$

(iv) From (ii) we get $BP=CP$. So $P$ is the midpoint of $BC$. Also $AP$ is an angle-bisector of the top vertex and it joins the vertex line to the midpoint of the base; in such symmetric isosceles configuration the line of symmetry through the vertex is perpendicular to the base. Concretely, triangles $ABP$ and $ACP$ are congruent and give $\mathrm{\angle }BPA=\mathrm{\angle }APC$. Since $\mathrm{\angle }BPA+\mathrm{\angle }APC={180}^{\circ }$ and the two are equal, each is ${90}^{\circ }$. Hence $AP\perp BC$ and it bisects $BC$. So $AP$ is the perpendicular bisector of $BC$. ✔

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Q2.

Statement: AD is an altitude of an isosceles triangle ABC with $AB=AC$. Prove: (i) AD bisects BC. (ii) AD bisects ∠A.

Solution.

Given $AB=AC$ and $AD\perp BC$. Consider triangles $ABD$ and $ACD$.

• $AB=AC$ (given),

• $AD=AD$ (common),

• $\mathrm{\angle }ADB=\mathrm{\angle }ADC={90}^{\circ }$.

So by RHS (right-angle, hypotenuse, side) congruence, $\mathrm{△}ABD\cong \mathrm{△}ACD$. From congruence:

(i) Corresponding parts give $BD=DC$, so $AD$ bisects $BC$.

(ii) Corresponding angles at A give $\mathrm{\angle }BAD=\mathrm{\angle }DAC$, so $AD$ bisects $\mathrm{\angle }A$

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Q3.

Statement: $AB=PQ,BC=QR$ and medians $AM$ and $PN$ satisfy $AM=PN$ (where $M$is midpoint of $BC$ and $N$ of $QR$). Prove: (i) $\mathrm{△}ABM\cong \mathrm{△}PQN$ (ii) $\mathrm{△}ABC\cong \mathrm{△}PQR$

Solution.

(i) Since $M$ and $N$ are midpoints,

$$BM=\frac{1}{2}BC,QN=\frac{1}{2}QR\mathrm{}$$

Given $BC=QR$ so $BM=QN$. Now compare triangles $ABM$ and $PQN$:

• $AB=PQ$ (given),

• $BM=QN$ (just shown),

• $AM=PN$ (given).

Thus by SSS, $\mathrm{△}ABM\cong \mathrm{△}PQN$

(ii) From (i) corresponding angles at $B$ and $Q$ are equal: $\mathrm{\angle }ABM=\mathrm{\angle }PQN$. In the full triangles $ABC$ and $PQR$ we have:

• $AB=PQ$ (given),

• $BC=QR$ (given),

• the included angle $\mathrm{\angle }ABC=\mathrm{\angle }PQR$ (because $\mathrm{\angle }ABM=\mathrm{\angle }PQN$ and $M,N$ lie on $BC,QR$ respectively giving the same full angle).
  Hence by SAS, $\mathrm{△}ABC\cong \mathrm{△}PQR$

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Q4.

Statement: BE and CF are two equal altitudes of triangle ABC (so $BE\perp AC$, $CF\perp AB$, and $BE=CF$). Using RHS, prove ABC is isosceles.

Solution.

Look at right triangles $ABE$ and $ACF$.

• $\mathrm{\angle }AEB=\mathrm{\angle }AFC={90}^{\circ }$,

• $BE=CF$ (given),

• $\mathrm{\angle }BAE=\mathrm{\angle }CAF$ — actually the angle at A is common to both triangles.

We can use AAS or view each as right triangles with equal leg and equal acute angle. Using RHS-type reasoning: the right triangles have their hypotenuses $AB$ and $AC$ as the sides opposite the right angles, and the equal legs $BE$and $CF$ correspond. From the congruence (RHS or AAS), $\mathrm{△}ABE\cong \mathrm{△}ACF$. Therefore corresponding hypotenuses are equal: $AB=AC$. So $ABC$ is isosceles. ✔

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Q5.

Statement: ABC is isosceles with $AB=AC$. Draw $AP\perp BC$ (P foot on BC). Prove $\mathrm{\angle }B=\mathrm{\angle }C$.

Solution.

Consider triangles $ABP$ and $ACP$. They are right-angled at $P$ (since $AP\perp BC$). Also:

• $AB=AC$ (given),

• $AP=AP$ (common),

• each has a right angle.

So by RHS congruence, $\mathrm{△}ABP\cong \mathrm{△}ACP$. Hence corresponding angles give $\mathrm{\angle }B=\mathrm{\angle }C$

(This also shows the altitude from the apex in an isosceles triangle is simultaneously an angle-bisector and perpendicular bisector of the base.)