Since DP AB and DQ AC, both BPD and CQD are right angles (90^°). Point D is given as the midpoint of BC, which means BD = CD. Additionally, it is given that DP = DQ.

(b) In the given figure, $D$ is the midpoint of $BC$ , $DP \perp AB$ , $DQ \perp AC$ and $DP = DQ$ . (i) Is $\triangle BPD \cong \triangle CQD$ ? Give reason. (ii) Is $\angle B = \angle C$ ? Justify your answer. Diagram Details: A triangle $ABC$ with base $BC$ and apex $A$ . Point $D$ lies on side $BC$ . A segment $DP$ is drawn from $D$ to side $AB$ such that $DP \perp AB$ with a right-angle symbol indicated at $P$ . A segment $DQ$ is drawn from $D$ to side $AC$ such that $DQ \perp AC$ with a right-angle symbol indicated at $Q$ . Segments $BD$ , $CD$ , $DP$ , and $DQ$ are shown.

Answer: (i) Yes, $\triangle BPD \cong \triangle CQD$ by the RHS congruence criterion. (ii) Yes, $\angle B = \angle C$ by CPCTC (Corresponding Parts of Congruent Triangles are Congruent).

Step-by-step solution

Step 1: Identify given relations for triangles BPD and CQD

Since $DP \perp AB$ and $DQ \perp AC$ , both $\angle BPD$ and $\angle CQD$ are right angles ( $90^\circ$ ). Point $D$ is given as the midpoint of $BC$ , which means $BD = CD$ . Additionally, it is given that $DP = DQ$ .

Step 2: Prove congruence of triangle BPD and triangle CQD using RHS

In right-angled triangles $BPD$ and $CQD$ , the hypotenuses $BD$ and $CD$ are equal, and one corresponding leg $DP$ equals $DQ$ . Therefore, by the Right angle-Hypotenuse-Side (RHS) congruence criterion, $\triangle BPD \cong \triangle CQD$ .

Step 3: Justify equality of angle B and angle C

Since $\triangle BPD$ is congruent to $\triangle CQD$ , their corresponding angles are equal by CPCTC (Corresponding Parts of Congruent Triangles are Congruent). Therefore, $\angle B$ equals $\angle C$ .

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