Question

Hence, $\angle AEB={90}^{\circ }$
Say true or false.

• A. True
• B. False
• C. Ambiguous
  
• D. Data insufficient
  

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 $△ABE$,
$\angle ABE+\angle BAE+\angle AEB={180}^o$ (Sum of angles of triangle)
$\angle ABE+\angle BAE+\angle BED+\angle DEA={180}^o$
$x+x+y+y={180}^o$
$x+y={90}^o$
$\angle DEB+\angle DEA={90}^o$
$\angle AEB={90}^{\circ }$