Question

In the triangle ABC AB=AC. side BC is produced to D and AE is perpendicular to BC. prove that AD square -AC square =BC. CD ​

Answer 1

$\textbf{Pythagors theorem:}$

$\text{In a right angled square on the hypotenuse is equal}$

$\text{to sum of the squares on the other two sides.}$

$\textbf{Given:}$

$\text{In isoceles \triangle ABC, AB=AC}$

$\textbf{To prove:}$

$AD^2-AC^2=BD.CD$

$\text{Since AE \perp BC, BE=EC}$.....(1)

$\text{In right \triangle AED, apply pythagoras theorem}$

$AD^2=AE^2+ED^2$ ........(2)

$\text{In right \triangle AEC, apply pythagoras theorem}$

$AC^2=AE^2+EC^2$ ........(3)

$\text{(2)-(3) gives}$

$AD^2-AC^2=(AE^2+ED^2)-(AE^2+EC^2)$

$AD^2-AC^2=ED^2-EC^2$

$\text{Using}$

$\boxed{\bf\,a^2-b^2=(a+b)(a-b)}$

$AD^2-AC^2=(ED+EC)(ED-EC)$

$AD^2-AC^2=(ED+EC)(CD)$

$AD^2-AC^2=(ED+BE)(CD)$ (Using (1))

$\implies\boxed{\bf\,AD^2-AC^2=(BD)(CD)}$