Q1(iii):

$\triangle ABC$ and $\triangle DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig. 7.39). If $AD$ is extended to intersect $BC$ at $P$, show that (iii) $AP$ bisects $\angle A$ as well as $\angle D$.

Step 1: Establishing Given Information and Geometric Constraints

Given that $\triangle ABC$ and $\triangle DBC$ are isosceles triangles sharing base $BC$, we establish the following equalities based on the definition of isosceles triangles [Property of Isosceles Triangles]:

• In $\triangle ABC$, $AB = AC$.
• In $\triangle DBC$, $DB = DC$.

To prove that $AP$ bisects $\angle A$ (i.e., $\angle BAP = \angle CAP$) and $\angle D$ (i.e., $\angle BDP = \angle CDP$), we must first verify the congruence of the constituent triangles formed by the line $AP$.

Step 2: Proving Triangle Congruence ($\triangle ABD \cong \triangle ACD$)

We compare $\triangle ABD$ and $\triangle ACD$:

1. $AB = AC$ [Given in $\triangle ABC$].
2. $BD = CD$ [Given in $\triangle DBC$].
3. $AD = AD$ [Common side].

By the SSS (Side-Side-Side) Congruence Criterion, $\triangle ABD \cong \triangle ACD$. Consequently, by CPCT (Corresponding Parts of Congruent Triangles), $\angle BAD = \angle CAD$. This confirms that $AP$ bisects $\angle A$.

Step 3: Proving Congruence to Establish Bisector Property at $D$

To evaluate the bisection of $\angle D$, we consider $\triangle BDP$ and $\triangle CDP$:

• $BD = CD$ [Given].
• $DP = DP$ [Common side].
• $\angle BDP = \angle CDP$ [To be determined].

First, we note that from $\triangle ABD \cong \triangle ACD$, it follows that $\angle ADB = \angle ADC$ [CPCT]. Since $\angle BDP$ and $\angle CDP$ are supplementary to the angles $\angle ADB$ and $\angle ADC$ respectively on the line $AD$, or more precisely, since they represent the angles formed by the transversal $AP$ through the isosceles triangle $DBC$, we look at $\triangle BDP$ and $\triangle CDP$:

In $\triangle BDP$ and $\triangle CDP$:

• $BD = CD$ [Given].
• $BP = CP$ [Derived from $\triangle ABP \cong \triangle ACP$ via SAS, where $AB=AC, \angle BAP=\angle CAP, AP=AP$].
• $DP = DP$ [Common side].

By SSS Congruence Criterion, $\triangle BDP \cong \triangle CDP$. Therefore, $\angle BDP = \angle CDP$ [CPCT].

Step 4: Visualizing the Geometry

Step 5: Final Deduction

Since $\angle BAP = \angle CAP$ (from $\triangle ABP \cong \triangle ACP$) and $\angle BDP = \angle CDP$ (from $\triangle BDP \cong \triangle CDP$), it is logically concluded that $AP$ acts as the angle bisector for both $\angle A$ and $\angle D$.

Final Solution: $AP$ bisects $\angle A$ as well as $\angle D$ due to the congruent triangle pairs $\triangle ABP \cong \triangle ACP$ and $\triangle BDP \cong \triangle CDP$ [Q.E.D.].