Question

In the isosceles $△ABC$, $AB=AC$, $BC=18cm$, $AD\perp BC$, $AD=12cm$, $BC$ is produced to '$E$' and $AE=20cm$. Prove that $\angle BAE={90}^o$.

Answer 1

In $△ABD$, $\angle D={90}^0$, $BD=9$ cm and $AD=12$ cm.

Using pythagoras theorem, we have

$A{B}^2=A{D}^2+B{D}^2\Rightarrow A{B}^2=9^2+{12}^2\Rightarrow A{B}^2=81+144\Rightarrow A{B}^2=225\Rightarrow AB=\sqrt{225}\Rightarrow AB=\sqrt{{15}^2}\Rightarrow AB=15$

In $△ADE$, $\angle D={90}^0$, $AE=20$ cm and $AD=12$ cm.

Again using pythagoras theorem, we have

$A{E}^2=A{D}^2+D{E}^2\Rightarrow {20}^2={12}^2+D{E}^2\Rightarrow 400=144+D{E}^2\Rightarrow D{E}^2=400-144\Rightarrow D{E}^2=256\Rightarrow DE=\sqrt{256}\Rightarrow DE=\sqrt{{16}^2}\Rightarrow DE=16$

From the figure, we observe that $BE=BD+DE$, therefore,

$BE=BD+DE=9+16=25$

Now, in $△ABE$, consider $A{B}^2+A{E}^2$ that is:

$A{B}^2+A{E}^2={15}^2+{20}^2=225+400=625=(25{)}^2=B{E}^2$

Since $A{B}^2+A{E}^2=B{E}^2$, therefore $△ABE$ is a right angled triangle.

Hence, $\angle BAE={90}^0$.