## AB is the diameter of the circle. DE is perpendicular on AB. Arc AC=arc DB To prove angle ABC=angle BDE

Apr 14, 2018

Given: AB is the diameter, arc AC= arcDB, $D E \bot A B$

RTP: $\angle A B C = \angle B D E$

Construction: . $A , D$ are joined.

Now in $\Delta A B D \mathmr{and} \Delta B E D$

$\angle A D B = \angle B E D = {90}^{\circ}$ as $\angle A D B$ is semicircular and $D E \bot A B$ is drawn. $\angle D B E = \angle A B D$ common

Hence remaining $\textcolor{red}{\angle B A D = \angle B D E \ldots \ldots \left[1\right]}$

So $\Delta A B D \mathmr{and} \Delta B E D$ are SIMILAR

Again in $\Delta A B C \mathmr{and} \Delta A B D$

$A C = B D$ as it is given arc AC= arc DB

$\angle A C B \mathmr{and} \angle A B D = {90}^{\circ}$ as they are semicircular angles.
and $A B$ hypotenuse is common.

So $\Delta A B C \cong \Delta A B D$

Hence $\textcolor{red}{\angle B A D = \angle A B C \ldots \ldots \left[2\right]}$

Combining [1] and [2] we get $\textcolor{m a \ge n t a}{\angle A B C = \angle B D E}$