ABC and DBC are two isosceles triangles on the same side of BC. Prove that :

(i) DA (or AD) produced bisects BC at right angle.

(ii) ∠BDA = ∠CDA.

Let ABC and DBC be two isosceles triangle with AB = AC and BD = CD.

(i) Let DA produced intersects BC at point L.

In △ ABL and △ ACL,

⇒ ∠ABL = ∠ACL (Since, AB = AC and angles opposite to equal sides are equal)

⇒ AB = AC (Given)

⇒ AL = AL (Common side)

∴ △ ABL ≅ △ ∠ACL (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠ALB = ∠ALC = x (let) and BL = CL.

From figure,

BLC is a straight line.

∴ ∠ALB + ∠ALC = 180°

⇒ x + x = 180°

⇒ 2x = 180°

⇒ x = $\dfrac{180°}{2}$ = 90°.

⇒ ∠ALB = ∠ALC = 90°.

Hence, proved that DA (or AD) produced bisects BC at right angle.

(ii) In △ BDL and △ CDL,

⇒ BD = CD (Given)

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

⇒ DL = DL (Common side)

∴ △ BDL ≅ △ ∠CDL (By S.A.S. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ ∠BDL = ∠CDL

From figure,

⇒ ∠BDL = ∠BDA and ∠CDL = ∠CDA

⇒ ∠BDA = ∠CDA.

Hence, proved that ∠BDA = ∠CDA.