Question

In the figure given alongside AB is parallel to CD and O is the center of the circle. If $\angle ADC=25,$ then find $\angle AEB.$

Answer 1

In the given figure join $AC$ and $BD$ then $\angle CAD=\angle CBD={90}^o$

$\because$ Angle is formed by diameter.

Also $AB||CD$

$\therefore \angle BAD=\angle ADC={25}^o$

$\angle BAC=\angle BAD+\angle CAD=25+90={115}^o$

$\therefore \angle ADB={180}^o-25-\angle BAC$$={40}^o$

Also, $\angle AEB=\angle ADB={40}^o$

Angle subtended by same chord on the circle are equal.