Question

ACB is tangent to a circle at C. CD and CE are chords such that $\angle ACE>\angle ACD$. If $\angle ACD=\angle BCE={50}^{\circ }$, then which is correct answer?

• A. CD = DE
• B. ED is not parallel to AB
• C. ED passes through the centre of the circle
• D. $△$CDE is a right angled triangle

Answer 1

The correct option is A CD = DE

Join DE, then
$\angle DEC=\angle ACD={50}^{\circ }$
[Angles in the alt. segments]
$\angle EDC=\angle BCE={50}^{\circ }$
[Angles in the alt. segments]
But $\angle ACD=\angle BCE$ (Given)
$\therefore \angle DEC=\angle EDC$
In $\Delta CDE,$
$\angle DEC=\angle EDC$ (Proved)
$\therefore$ CD = CE
(Sides opposite to equal $\angle s$ are equal)
$\therefore$ (A) is the correct answer.