Question

Chord ED is parallel to the diameter AC of the circle. Given ∠CBE=55°, calculate ∠DEC.

Answer 1

Consider the arc CDE.

The angles in the same segment are:

$\angle CBEand\angle CAE$
$\therefore \angle CAE=\angle CBE$
$\Rightarrow \angle CAE={55}^{\circ }[\angle CBE={55}^{\circ }]$
Since AC is the diameter of the circle and the angle in a semi - circle is a right angle
Therefore,
$\angle AEC={90}^{\circ }$.

Now, in $\Delta ACE$, we have
$\angle ACE+\angle AEC+\angle CAE={180}^{\circ }$
$\Rightarrow \angle ACE+{90}^{\circ }+{55}^{\circ }={180}^{\circ }$
$\Rightarrow \angle ACE={35}^{\circ }$.


But $\angle DEC$ and $\angle ACE$ are alternate angles, because $AC||DE$.
$\therefore \angle DEC=\angle ACE={35}^{\circ }$.