Question

In $\Delta ABC,\angle ABC={90}^o$ and $BD\perp AC$, then prove that $\Delta ABD\sim \Delta BCD$.

Answer 1

In $\Delta ABC\Rightarrow \alpha +\theta +90=180$

$\alpha +\theta ={90}^{\circ }$ ____ (1)

In $\Delta BDC\Rightarrow 90+\theta +x={180}^{\circ }$

$\theta +x={90}^{\circ }$ _____ (2)

From (1) & (2)

$\theta +x=\theta +\alpha$

$x=\alpha$

$\Delta ABC$ & $\Delta ADB\Rightarrow$ by $y=\theta$

$\therefore \Delta ABD\sim \Delta BCD$ [3 angle are equal & one side in common]