In the given figure, AB = AC; D is the mid-point of BC; DP ⊥ BA and DQ ⊥ CA. Prove that:

(i) DP = DQ

(ii) AP = AQ

(iii) AD bisects ∠A

(i) In △ABC,

⇒ AB = AC (Given)

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

In △PDB and △QDC,

⇒ BD = CD (D is mid-point of BC)

⇒ ∠B = ∠C (Proved above)

⇒ ∠P = ∠Q (Both equal to 90°)

∴ △PDB ≅ △QDC (By A.A.S axiom)

⇒ DP = DQ (Corresponding parts of congruent triangles are equal)

Hence, proved that DP = DQ.

(ii) Since, △PDB ≅ △QDC

∴ BP = QC = y (let) (Corresponding parts of congruent triangles are equal)

⇒ AB = AC = x (let)

From figure,

⇒ AP = AB - BP = x - y …(1)

⇒ AQ = AC - QC = x - y …(2)

From eq.(1) and (2) we have :

∴ AP = AQ.

Hence, proved that AP = AQ.

(iii) In △ABD and △ACD,

⇒ AB = AC (Given)

⇒ BD = CD (Given)

⇒ AD = AD (Common side)

∴ △ABD ≅ △ACD (By S.S.S axiom)

⇒ ∠BAD = ∠CAD (Corresponding parts of congruent triangles are equal)

Hence, proved that AD bisects ∠A.