In the adjoining figure, AB=AC and BD=DC. Prove that and hence sh… - Vidyadip

Question

In the adjoining figure, $\text{AB}=\text{AC}$ and $\text{BD}=\text{DC}.$ Prove that $\triangle\text{ADB}\cong\triangle\text{ADC}$ and hence show that.

$\angle\text{ADB}=\angle\text{ADC}=90^\circ$
$\angle\text{BAD}=\angle\text{CAD}$

✓

Answer

In $\triangle\text{ABD}$ and $\triangle\text{ACD},$
$\text{AD}=\text{AD}$ (common)
$\text{AB}=\text{AC}$ (given)
$\triangle\text{ABD}\cong{}\triangle\text{ADC}$ (SSS condition)
$\angle\text{BAD}=\angle\text{CAD}$ (c.p.c.t.)
and $\angle\text{ADB}=\angle\text{ADC}$ (c.p.c.t.)
But $\angle\text{ADB}+\angle\text{ADC}=180^\circ$ (Linear pair)
$\angle\text{ADB}=\angle\text{ADC}=90^\circ$
Hence proved.

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