Question

In the given figure, chords AB and CD of a circle
intersect at E. Prove that∆ ADE and ∆CBE are
equiangular.​

Answer 1

Two chords AB and CD intersect each other at the point E,

we have to prove:---

Triangle ADE and Triangle CBE are equiangular,

In Triangle ADE and Triangle CBE , we get following things:---

1), Angle AED = angle CEB ( vertically opposite),

2), angle EAD = angle CEB, (angles In the same segments are equal)

3) angle ADE = angle EBC,( angles in same segments are equal)

Therefore Triangle ADE and Triangle CBE are equiangular,

Hence proved,

Answer 2

ΔADE and ΔCBE are equiangular.

Step-by-step explanation:

Given,two triangles ΔAED and ΔCEB where chord AB and CD intersect at E.

To prove, ΔADE and ΔCBE are equiangular.

From the figure ,

∠AED = ∠CEB (Vertically opposite angles)

∠A = ∠ C   (Since ∠A and ∠C are the angle subtended on the same arc)

Similarly ∠D = ∠B

As all the angle are equal.

∴, ΔAED ≅ ΔCEB

Also,ΔADE and ΔCBE are equiangular.

Hence Proved

*Refer this link for figure

https://brainly.in/question/2017861