Question

In figure, ABC is a right triangle right angled at A. D lies on BA produced and DE$\perp$BC, intersecting AC at F. If $\angle AFE={130}^o$, find $\angle ABC$.

Answer 1

Given: $\angle BAC={90}^{\circ },DE\perp BC,\angle AFE={130}^{\circ }$

To find: $\angle ABC$

Solution: Clearly, $\angle BAC+\angle FAD={180}^{\circ }$[Linear pair]

$\Rightarrow {90}^{\circ }+\angle FAD={180}^{\circ }$

$\Rightarrow \angle FAD={180}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle FAD={90}^{\circ }$

Now, By Exterior angle sum property of triangle, we have

$\angle AFE=\angle FAD+\angle ADF$

$\Rightarrow {130}^{\circ }={90}^{\circ }+\angle ADF$

$\Rightarrow \angle ADF={130}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle ADF={40}^{\circ }$

$\Rightarrow \angle BDE=\angle ADF={40}^{\circ }$

Now, In $△BDE$ by angle sum property of triangle, we have

$\angle BDE+\angle BED+\angle EBD={180}^{\circ }[\because DE\perp BC\Rightarrow \angle BED={90}^{\circ }]$

$\Rightarrow {40}^{\circ }+{90}^{\circ }+\angle EBD={180}^{\circ }$

$\Rightarrow \angle EBD={180}^{\circ }-{40}^{\circ }-{90}^{\circ }$

$\Rightarrow \angle EBD={50}^{\circ }$

$\Rightarrow \angle EBD=\angle ABC={50}^{\circ }$