Question

In the given figure, two chords AC and BD of a circle intersect at E. If arc AB = arc CD,
Then which of the following is true?

• A. BE = AE .
• B. AE=AC and BE=BD
• C. BE = EC and AE = ED.
• D. none of the above

Answer 1

The correct option is C BE = EC and AE = ED.

Given- The two chords AC & BD of a circle intersect at E. $arcAPB=arcCQD$.
To find out- which of the given options is true.
Solution- We join AB, CD and AD. $arcAPB=arcCQD⟹AB=CD.......\left(i\right)$ (because equal arcs of a circle contain equal chords). Now the chord AD subtends $\angle ABD=\angle ACD$ to the circumference of the given circle at B & C respectively.
$\therefore \angle ABD\&\angle ACD$ (because the angles, subtended by a chord of a circle to different points of the circumference of the same circle, are equal)........(ii)
Also $\angle AEB=DEC$ (vertically opposite angles).
$\therefore$ The third angles i.e $\angle BAE=\angle CDE..........\left(iii\right)$
So from (i), (ii) & (iii) we conclude that $\Delta AEB\cong \Delta CED$
$⟹BE=ECandAE=ED$.
Ans- Option C.