Question

In the given figure, triangle ABC is a right angle triangle with $\angle B={90}^{\circ }$ and D is mid point of side BC. Prove that:
${AC}^2={AD}^2+3{CD}^2$

Answer 1

Pythagoras theorem states that in a right angled triangle, the square on the hypotenuse is equal to the sum of of the squares on the remaining two sides.
In triangle $ABC,\angle B={90}^o$ and $D$ is the mid-point of $BC$. JOin $AD$. Therefore, $BD=DC$
First, we consider the $△ADB$ and applying
Pythagoras theorem we get,
$A{D}^2=A{B}^2+B{D}^2$
$A{B}^2=A{D}^2-B{D}^2...\left(i\right)$
Similarly, we get from rt. angle triangles $ABC$ we get,
$A{C}^2=A{B}^2+B{C}^2$
$A{B}^2=A{C}^2-B{C}^2...\left(ii\right)$
Grom $\left(i\right)$ and $\left(ii\right)$,
$A{C}^2-B{C}^2=A{D}^2-B{D}^2$
$A{C}^2=A{D}^2-B{D}^2+B{C}^2$
$A{C}^2=A{D}^2-B{D}^2+B{C}^2$
$A{C}^2=A{D}^2-C{D}^2+4C{D}^2[BD=CD=\frac{1}{2}BC]$
$A{C}^2=A{D}^2+3C{D}^2$
Hence Proved.