Question

The chord ED is parallel to the diameter AC as shown in the figure. The magnitude of $\angle CED$ is equal to:

• A. ${30}^{\circ }$
• B. ${40}^{\circ }$
• C. ${50}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is B ${40}^{\circ }$

Here AC is the diameter. but the diameter passes through the chord BE. We know that a $\perp$ line draw from the center to the chord bisects the chord.
$\therefore BP=PE$ by SAS congruence, $△CBP$ and $△CEP$ are congruent
$\therefore BC=EC$(CPCT)
$\angle CBE=\angle CEB$ (Angles opposite to equal sides)
$50°=50°$
But $\angle BPC=90°$ and it is also an interior angle of $△CEP$
$\angle BPC=\angle ECP+\angle CEP90=\angle 2+50\angle 2=40$
Now, $AC||ED$
$\therefore \angle CED=\angle 2=40°$(Alternate interior angle)