In the given figure, AB = AC and ∠DBC = ∠ECB = 90°.

Prove that :

(i) BD = CE

(ii) AD = AE

(i) In △ ABC,

⇒ AB = AC (Given)

⇒ ∠ABC = ∠ACB (Angles opposite to equal sides are equal) …….(1)

From figure,

⇒ ∠DBC = ∠ECB (Both equal to 90°) …….(2)

Subtracting equation (1) from (2), we get :

⇒ ∠DBC - ∠ABC = ∠ECB - ∠ACB

⇒ ∠DBA = ∠ECA ………(3)

In △ DBA and △ ECA,

⇒ ∠DBA = ∠ECA (Proved above)

⇒ AB = AC (Given)

⇒ ∠DAB = ∠EAC (Vertically opposite angles are equal)

∴ △ DBA ≅ △ ∠ECA (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ BD = CE.

Hence, proved that BD = CE.

(ii) Since,

△ DBA ≅ △ ∠ECA

∴ AD = AE (By C.P.C.T.C.)

Hence, proved that AD = AE.