Question

In the adjoining figure, E is a point on side CB produced of an isosceles ∆ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF.

Answer 1

In ∆ABC,

seg AB ≅ seg AC ---- [Given]

∠B ≅ ∠C ---- (i) [By isosceles triangle theorem]

In ∠ABD and ∠ECF,

∠ABD ≅ ∠ECF ---- [From (i)]

∠ADB ≅ ∠EFC ---- [Each is 90°]

 \(\therefore\) ∆ABD ~ ∆ECF ---- [By A–A test of similarity]