Question

In the given figure, $AD=DB=DE$. $\angle EAC=\angle FAC$ and $\angle F={90}^{\circ }$ Such that:$\angle AEB={85}^o$

• A. True
• B. False

Answer 1

The correct option is B False

In $△DBE$,

$DB=DE$

Hence, $\angle DBE=\angle DEB=x$ (I)
In $△DAE$,

$DA=DE$

Hence, $\angle DAE=\angle DEA=y$ (II)
Now, In $△ABE$,

$\angle ABE+\angle BAE+\angle AEB={180}^{\circ }$ (Sum of angles of triangle)

$\angle ABE+\angle BAE+\angle BED+\angle DEA={180}^{\circ }$

$x+x+y+y={180}^{\circ }$

$x+y={90}^{\circ }$

$\angle DEB+\angle DEA={90}^{\circ }$

$\angle AEB={90}^{\circ }$