Question

∆ABC right triangle , right angled at B. BD is perpendicular to AC . Prove that ∆ADB os is similar to ∆ABC.

Answer 1

Step-by-step explanation:

in ∆ABC and ∆ADB,

∠B = ∠ADB (∵BD⊥AC)

and ∠A = ∠BAD (common)

=> ΔABC ~ ΔADB (by AA criterion)