In the given figure, AB = AC. Prove that :

(i) DP = DQ

(ii) AP = AQ

(iii) AD bisects angle A

(i) In △ ABC,

⇒ AB = AC (Given)

∴ ∠B = ∠C (Angles opposite to equal sides are equal)

In △ PDB and △ QDC,

⇒ ∠P = ∠Q (Both equal to 90°)

⇒ ∠B = ∠C (Proved above)

⇒ BD = CD (Given)

∴ △ PDB ≅ △ QDC (By A.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ DP = DQ

Hence, proved that DP = DQ.

(ii) Since, △ PDB ≅ △ QDC

∴ BP = QC = y (let) [By C.P.C.T.C.]

⇒ AB = AC = x (let)

From figure,

⇒ AP = AB - BP = x - y

⇒ AQ = AC - QC = x - y

∴ AP = AQ.

Hence, proved that AP = AQ.

(iii) Join AD.

In △ ABD and △ ACD,

⇒ AB = AC (Proved above)

⇒ BD = CD (Given)

⇒ AD = AD (Common side)

∴ △ ABD ≅ △ ACD (By S.S.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠BAD = ∠CAD.

Hence, proved that AD bisects angle A.