Question
In the figure given below, if $\text{AC}=\text{AD}=\text{CD}=\text{BD};$ find angle $\text{ABC}$.
✓
Answer
In $\triangle ACD,$
$\text{AC} =\text{ AD} = \text{CD} ......[$Given$]$
Hence, $\text{ACD}$ is an equilateral triangle.
$\therefore \angle ACD = \angle CDA = \angle CAD = 60^\circ $
$\angle CDA = \angle DAB + \angle ABD ........[$Ext angle is equal to the sum of opp. int. angles$]$
But,
$\angle DAB = \angle ABD ......[$Given$: \text{AD} =\text{ DB}]$
$\therefore \angle ABD + \angle ABD = \angle CDA$
$\Rightarrow 2\angle ABD = 60^\circ $
$\Rightarrow \angle ABD = \angle ABC = 30^\circ $
$\text{AC} =\text{ AD} = \text{CD} ......[$Given$]$
Hence, $\text{ACD}$ is an equilateral triangle.
$\therefore \angle ACD = \angle CDA = \angle CAD = 60^\circ $
$\angle CDA = \angle DAB + \angle ABD ........[$Ext angle is equal to the sum of opp. int. angles$]$
But,
$\angle DAB = \angle ABD ......[$Given$: \text{AD} =\text{ DB}]$
$\therefore \angle ABD + \angle ABD = \angle CDA$
$\Rightarrow 2\angle ABD = 60^\circ $
$\Rightarrow \angle ABD = \angle ABC = 30^\circ $
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