Question

In $\Delta$ ABC, $\angle B={90}^0$ and D is midpoint of BC. Prove that : $A{C}^2=A{D}^2+3C{D}^2$

Answer 1

Given: $△ABC$, $\angle B={90}^{\circ }$, D is mid point of BC.
In $△ABC$,
$A{C}^2=B{C}^2+A{B}^2$
$A{C}^2=A{B}^2+(2CD{)}^2$ (D is mid point of BC)
$A{C}^2=A{B}^2+4C{D}^2$... (I)
In $△ABD$,
$A{D}^2=A{B}^2+B{D}^2$
$A{D}^2=A{B}^2+C{D}^2$ (BD = CD)...(II)
From I and II, equate $A{B}^2$
$A{C}^2-4C{D}^2=A{D}^2-C{D}^2$
$A{C}^2=A{D}^2+3C{D}^2$