Question
In the figure, $BC = CE$ and $\angle 1 = \angle 2.$ Prove that $\triangle GCB ≅ \triangle DCE.$
✓
Answer
In $\triangle GCB$ and $\triangle DCE$ and
$\angle 1 + \angle GBC = \angle 2 + \angle DEC = 180^\circ $
$\angle 1 = \angle 2 =$
$\Rightarrow \angle GBC = \angle DEC$
$BC = CE$
$\angle GCB = \angle DCE = ...($vertically opposite angles$)$
Therefore,
$\triangle GCB ≅ \triangle DCE ....(\text{ASA}$ criteria$)$.
$\angle 1 + \angle GBC = \angle 2 + \angle DEC = 180^\circ $
$\angle 1 = \angle 2 =$
$\Rightarrow \angle GBC = \angle DEC$
$BC = CE$
$\angle GCB = \angle DCE = ...($vertically opposite angles$)$
Therefore,
$\triangle GCB ≅ \triangle DCE ....(\text{ASA}$ criteria$)$.
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\begin{array}{c}
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\end{array}
$
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