In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB. Prove that :

(i) BD = CD

(ii) ED = EF

(i) In △ ABD and △ ACD,

⇒ AD = AD (Common side)

⇒ AB = AC (Given)

⇒ ∠ADB = ∠ADC (Since, AD is perpendicular to BC)

∴ ∆ ABD ≅ ∆ ACD (By R.H.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ BD = CD.

Hence, proved that BD = CD.

(ii) In △ EBD and △ EBF,

⇒ EB = EB (Common side)

⇒ ∠EBF = ∠EBD (Since, BE bisects angle B)

⇒ ∠EFB = ∠EDB (Both equal to 90°)

∴ ∆ EBD ≅ ∆ EBF (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangles are equal.

∴ ED = EF.

Hence, proved that ED = EF.