Question

In the quadrilateral $ABCD$, $\angle ADC={90}^o$, $AB=9cm$, $BC=AD=6cm$ and $CD=3cm$. Prove that $\angle ACB={90}^o$

Answer 1

In $△ADC$, $\angle D={90}^0$, $AD=6$ cm and $CD=3$ cm.

Using pythagoras theorem, we have

$A{C}^2=A{D}^2+C{D}^2\Rightarrow A{C}^2=6^2+3^2\Rightarrow A{C}^2=36+9\Rightarrow A{C}^2=45\Rightarrow AC=\sqrt{45}\Rightarrow AC=\sqrt{3^2\times 5}\Rightarrow AC=3\sqrt{5}$

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

$A{C}^2+B{C}^2=(3\sqrt{5}{)}^2+6^2=(9\times 5)+36=45+36=81=(9{)}^2=A{B}^2$

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

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