Question

In a cyclic quadrilateral A B C D, E is a point on AC such that $\angle ADE=\angle BDC$.

Prove that(i) AD. BC = AE . DB and (ii)AB. DE = CE . DA

Answer 1

In a cyclic quadrilateral, the vertices are on the circumference of a circle. Also, the angle between a side and a diagonal is equal to the angle between the opposite side and the other diagonal.

Given ∠ADE = ∠CDB = x (let).

ABCD is a Cyclic Quadrilateral

=> ∠CAD = ∠CBD = y and ∠CDB = ∠CAB = x

=> ∠ABD = ACD = w

Also, ∠CDE = x + z = ∠ADB

Compare ΔAED and ΔBCD. The angles are same and hence are similar triangles. Ratios of corresponding sides are:

=> AD/ BD = AE / BC

=> AD * BC = AE * BD proved.

Now, compare ΔABD and ΔCDE. Two angles x+z and w are equal in both. They are similar triangles. Hence,

=> AB / DA = CE / DE

=> AB * DE = CE * DA proved.