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Question

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 $$\Delta AEB$$ is isosceles.
1072142_55f224d12834492a8735a476d35fd1b1.png


Solution

$$AB$$ and $$CD$$ are parallel chords of circle. Line $$BD$$ and $$AC$$ intersect at point $$E$$.

As we know that in a  cyclic quadrilateral, the exterior angle is equal to opposite interior angle then

Exterior angle $$\angle CDE=\angle A,$$                             ......(1)

Exterior angle $$\angle ECD=\angle B$$                             ......(2)

Now, $$AB\parallel CD$$ then

$$\angle A=\angle ECD$$    when $$AE$$ is tranversal,

Hence,
$$\angle A=\angle B.$$                     [By equation (2)]

Also, we know in a triangle, opposite sides of equal angles are equal then 
$$AE=BE$$                     (in $$\triangle AEB$$ )

Hence, $$\triangle AEB$$ is isosceles.

Mathematics
RS Agarwal
Standard IX

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