Question

In the given figure, $AD=DB=DE$. $\angle EAC=\angle FAC$ and $\angle F={90}^{\circ }$ Such that:$△CEG$ is isosceles.

• A. True
• B. False

Answer 1

The correct option is A True

In $△DBE$,

$DB=BE$

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

$DA=AE$

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

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

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

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

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

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

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

Hence, $\angle AEB=\angle AEC={90}^{\circ }$
In $△AEF$,

Sum of angles$={180}^{\circ }$

$\angle AEF+\angle EAF+\angle EFA={180}^{\circ }$

$\angle AEF+\angle EAF+{90}^{\circ }={180}^{\circ }$

$\angle AEF={90}^{\circ }-\angle EAF$
We know, $\angle AEC={90}^{\circ }$

$\angle AEF+\angle FEC={90}^{\circ }$

${90}^{\circ }-\angle EAF+\angle FEC={90}^{\circ }$

or $\angle EAF=\angle FEC$

Now, In $△CEA$,

Sum of angles $={180}^{\circ }$

$\angle AEC+\angle EAC+\angle ACE={180}^{\circ }$

$\angle ACE={90}^{\circ }-\angle EAC$ ..(III)
Now, in $△GCE$

Sum of angles$={180}^{\circ }$

$\angle GCE+\angle GEC+\angle CGE={180}^{\circ }$

$({90}^{\circ }-\angle EAC)+2\angle EAC+\angle CGE={180}^{\circ }$

$\angle GCE={90}^{\circ }-\angle EAC$ ...(IV)

hence, from III and IV,

$\angle GCE=\angle ACE$

or $GCE$ is an Isosceles triangle.