Question

In the given figure, ∆ABC is an isosceles triangle with AB= AC and angle ABC = 50°. Find angle BDC and angle BEC.​

Answer 1

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Solution

Answer 2

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$\huge\mathfrak\purple{Solution:-}$

⏩ Answer:-

angle BDC = 80° and angle BEC = 100°

⏩ Step-by -step explanation:-

In ∆ABC, we have,

AB = AC

=> angle ACB = angle ABC

=> angle ACB = 50° (since, angle ABC = 50°)

Therefore,

angle BAC = 180° - (angle ABC + angle ACB)

=> angle BAC = 180° - (50° + 50° ) = 80°

Now,

Since angle BAC and angle BDC are angles in the same segment.

Therefore,

angle BDC = angle BAC

=> angle BDC = 80°

Now,

BDCE is a cyclic quadrilateral.

Therefore,

angle BDC + angle BEC = 180°

=> 80° + angle BEC = 180°

=> angle BEC = 100°

Hence,

⏩angle BDC = 80° and angle BEC = 100°

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