Question

In ∆ABC, given that AB = AC and BD ⊥ AC. Prove that BC² = 2 AC × CD.

Answer 1

Since ΔADB is right triangle right angled at D

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

Substitute `AB=AC`

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

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

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

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

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

Now, in ΔBDC, we have

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

Therefore, `2AC.DC = DC^2+BD^2`

`2AC.DC=BC^2`

Hence proved.