Question
In the figure $\triangle ABC$ is isosceles with $AB = AC.$ Prove that:$\angle ADE = \angle BCD$
✓
Answer
In $\triangle ADE,$
$\angle ADE = a ...($from $(i))$
In $\triangle BCD,$
$\angle BCE = b ...(AB = AC \Rightarrow \angle B = \angle C) .........(vii)$
$\angle BCD = \angle BCE - \angle DCE$
$\angle BCD = b - 2a ...($from $(vii)$ and $(ii))$
But $b = 3a ...($from $(vi))$
Therefore, $\angle BCD = 3a - 2a = a$
Hence, $\angle ADE = \angle BCD.$
$\angle ADE = a ...($from $(i))$
In $\triangle BCD,$
$\angle BCE = b ...(AB = AC \Rightarrow \angle B = \angle C) .........(vii)$
$\angle BCD = \angle BCE - \angle DCE$
$\angle BCD = b - 2a ...($from $(vii)$ and $(ii))$
But $b = 3a ...($from $(vi))$
Therefore, $\angle BCD = 3a - 2a = a$
Hence, $\angle ADE = \angle BCD.$
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