In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that △AEB is isosceles.
Open in App
Solution
ANSWER: AB and CD are two parallel chords of a circle. BDE and ACE are two straight lines that intersect at E. If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle. ∴ Exterior ∠EDC = ∠A ...(i) Exterior ∠DCE = ∠B ...(ii) Also, AB parallel to CD. Then, ∠EDC = ∠B (Corresponding angles) and ∠DCE = ∠A (Corresponding angles) ∴ ∠A = ∠B [From(i) amd (ii)] Hence, ΔAEB is isosceles.