Question
Two right$-$angled triangles $ABC$ and $ADC$ have the same base $AC$. If $BC = DC$, prove that $AC$ bisects $\angle BCD.$
✓
Answer
In $\triangle ABC$ and $\triangle ADC$
$\angle BAC = \angle DAC ...(90^\circ )$
$BC = DC$
$AC = AC ...($common$)$
Therefore, $\triangle ABC ≅ \triangle ADC ...(\text{SSA}$ criteria$)$
Hence, $\angle BCA = \angle DCA$
Thus, $AC$ bisects $\angle BCD.$
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