Question

plsss some1 solve this it's very imp. I've a test today nd m not able to solve anything.so plsss

Answer 1

The given quadrilateral is a cyclic quadrilateral as all its four vertices lie on the circle.
We know that in a cyclic quadrilateral, opposite sides are supplementary(sum of angles=180).
Therefore, <ABC=50°
Now, consider the two triangles.
given, BC=BE.
both are right triangles with an angle of 90°.
The triangles have a common side.
Hence, by RHS congruency criterion,
the triangles are congruent.
therefore, <ABC=<ABE.=50°
Hence, <ABC+<ABE=100°.


Hope this helps you......

Answer 2

hey there !!

=) ABCD is a cyclic quadrilateral so we know that from property of cyclic quadrilateral ..

=) /_ADC +/_CBA= 180°

=) 130° +/_CBA = 180°

=) /_CBA = 180 -130 = 50°

Here , given that CB = BE

=) we know that opposite side of ∆ are equal in isosceles ∆

◆from property of isosceles ∆

=) /_CBA = /_ABE = 50°

so /_ CBE = /_CBA + /_ABE

= 50 + 50 = 100°

* ,* /_ CBE = 100 ° ans//

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Hope it will help u