Question

In an isoceles triangle ,∆BCD in which BA is perpendicular to CD and A is the mid point of CD. Show that angle CBA = angle DBA. Above is the rough figure. Please answer it.​

Answer 1

Explanation:

A is the mid point of CD ⇒ given

∠A = ∠A ⇒ Each 90°

BA is ⊥ CD

So, AB bisects ∠B

∴ ∠CBA = ∠DBA