Question

In ∆ABC, if BD ⊥ AC and BC² = 2 AC . CD, then prove that AB = AC.

Answer 1

Since Δ ADB  is right triangle right angled at D

`AB^2=AD^2+BD^2`

In right Δ BDC, we have

`CD^2+BD^2=BC^2`

Sine `2AC.DC=BC^2`

`⇒DC^2+BD^2=2AC.DC `

`2AC.DC=AC^2-AC^2+DC^2+BD^2`

`AC^2=AC^2+DC^2-2AC.DC+BD^2`

`AC^2=(AC-DC)^2+BD^2`

`AC^2=AD^2+BD^2`

Now substitute `AD^2+BD^2=AB^2`

`AC^2=AB^2`

`AC=AB`