Question

In the given figure, in $∆$ABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD $\cong$seg BE

Answer 1

In △ADB and △BEA
∠ADB = ∠BEA = 90^∘ (Given)
BD = AE (Given)
AB = BA (Common)
By RHS test of congruency
△ADB ≅ △BEA
∴ AD = BE (corresponding angles of congruent triangles)
Hence proved.