In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF
Advertisement Remove all ads
Solution
It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ABD = ∠ECF
In ΔABD and ΔECF,
∠ADB = ∠EFC (Each 90°)
∠ABD = ∠ECF (Proved above)
∴ ΔABD ∼ ΔECF (By using AA similarity criterion)
Concept: Criteria for Similarity of Triangles
Is there an error in this question or solution?
Advertisement Remove all ads
APPEARS IN
Advertisement Remove all ads
Advertisement Remove all ads