Question

In $\Delta$ABC, D is a mid point of side AC such that BD$=\frac{1}{2}$AC, then$\angle$ABC is a right angle.

• A. True
• B. False

Answer 1

The correct option is A True

$\begin{array}{c}if\Delta ABCisrightanglethen,\\ A{C}^2=A{B}^2+B{C}^2...\left(1\right)\\ andin\Delta BCD\\ B{C}^2=B{D}^2+\frac{A{C}^2}{4}...\left(2\right)\\ In\Delta ABD\\ A{B}^2=B{D}^2+\frac{A{C}^2}{4}....\left(3\right)\\ Addingeq\left(1\right)and\left(2\right)\\ B{C}^2+A{B}^2=2B{D}^2+\frac{A{C}^2}{2}\\ fromeq\left(1\right)\\ A{C}^2=2B{D}^2+\frac{A{C}^2}{2}\\ A{C}^2-\frac{A{C}^2}{2}=2B{D}^2\\ \frac{A{C}^2}{2}=2B{D}^2\\ A{C}^2={\left(2BD\right)}^2\\ AC=2BD\\ so,itistrue\end{array}$