Question

In triangle ABC, if BD perpendicular to AC and BC2=2AC.CD, prove that AB=AC
(class 10)

Answer 1

hope it helps you......

Answer 2

Pythagoras Theorem is the Helping Hand

Step-by-step explanation:

Given,

BD is perpendicular to AC and $BC^2 = 2 AC \times CD$

We have,  

$BC^2 = 2 AC \times CD$

⇒ $BD^2 + CD^2 = 2 AC \times CD$  [Using pythagoras theorem in Δ BCD]

Again,  in Δ ABD by Pythagoras Theorem, we have

$AB^2 = BD^2 + AD^2$

⇒ $BD^2 = AB^2 - AD^2$

So, $AB^2 - AD^2 + CD^2 = 2 AC \times CD$

⇒ $AB^2 - (AC - CD)^2 + CD^2 = 2 AC \times CD$

⇒ $AB^2 - AC^2 - CD^2 + 2 AC \times CD + CD^2 = 2 AC \times CD$

⇒ $AB^2 - AC^2 = 0$

⇒ $AB^2 = AC^2$

⇒ AB = AC   [Hence proved]