Question

In the figure (ii) given below, AB is the diameter of the semicircle ABCDE with center O. If AE=ED and $\angle BCD={140}^{\circ }$, find $\angle AED$ and $\angle EBD$. Also prove that OE is parallel to BD.

Answer 1

Given:

AB is the diameter of semi-circle ABCDE with centre $O$. $AE=ED$

to find:

∠AED∠AED and ∠EBD

Solution:

Since, $\angle BCD={140}^o$.

In cyclic quadrilateral $EBCD$.

$\angle BCD+\angle BDE={180}^o$

$\Rightarrow {140}^o+\angle BED={180}^o$

$\Rightarrow \angle BDE={180}^o-{140}^o={40}^o$

But $\angle AEB={90}^o$ (angle in a semi-circle)

$\therefore AED=\angle AEB+\angle BED$

$={90}^o+{40}^o={130}^o$

Now in cycle quadrilateral $AEDB$

$\angle AED+\angle DBA={180}^o$

$\Rightarrow {130}^o+\angle DBA={180}^o$

$\Rightarrow \angle DBA={180}^o-{130}^o={50}^o$

$\because$ chord $AE=ED$ (given)

$\therefore \angle DBE=\angle EBA$

But $\angle DBE+\angle EBA={50}^o$

$\Rightarrow DBE+\angle DBE={50}^o$

$2\angle DBE={25}^o$

$\therefore \angle DBE={25}^o$

or

$\angle EBD={25}^o$

In $\Delta DEB,OE=OB$ (radii of the same circle)

therefore

$\angle OEB=\angle BBO=\angle DBE$

But these are alternate angles.

$\therefore OE||BD$ (Q.E.D)