Question

ABC is an isosceles triangle with AB=AC .D is a point on AC such that BC²=AC×AD .prove that triangle ABC is similar to triangle DBC​

Answer 1

Answer:

Answer

We have,

BC

2

=AC×CD and AB=AC

⇒ BC×BC=AC×CD and ∠B=∠C

⇒

AC

BC

=

BC

CD

and ∠B=∠C

⇒

CA

BC

=

CB

DC

and ∠B=∠C

So, by SAS-criterion of similarity, we have

△BCA∼△DCB

⇒

DC

BC

=

CB

CA

=

DB

BA

⇒

CB

CA

=

DB

BA

⇒

CA

BA

=

CB

DB

⇒ 1=

CB

DB

[∵ BA=CA]

⇒ DB=CB

⇒BD=BC [Hence proved