Question
From the following figure,
prove that $: \angle ACD =\angle CBE$
prove that $: \angle ACD =\angle CBE$
✓
Answer
In $\triangle ACB,$
$AC = AC ....[$Given$]$
$\therefore \angle ABC = \angle ACB ...(i) [$angles opposite to equal sides are equal$]$
$\angle ACD + \angle ACB = 180^\circ ......(ii) [DCB$ is a straight line$]$
$\angle ABC + \angle CBE = 180^\circ ….(iii)[ ABE$ is a straight line $]$
Equating $(ii)$ and $(iii)$
$\angle ACD + \angle ACB = \angle ABC + \angle CBE$
$\Rightarrow \angle ACD + \angle ACB = \angle ACB + \angle CBE ......[$From $(i)]$
$\Rightarrow \angle ACD = \angle CBE.$
$AC = AC ....[$Given$]$
$\therefore \angle ABC = \angle ACB ...(i) [$angles opposite to equal sides are equal$]$
$\angle ACD + \angle ACB = 180^\circ ......(ii) [DCB$ is a straight line$]$
$\angle ABC + \angle CBE = 180^\circ ….(iii)[ ABE$ is a straight line $]$
Equating $(ii)$ and $(iii)$
$\angle ACD + \angle ACB = \angle ABC + \angle CBE$
$\Rightarrow \angle ACD + \angle ACB = \angle ACB + \angle CBE ......[$From $(i)]$
$\Rightarrow \angle ACD = \angle CBE.$
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