Question
In the adjoining figure, $\angle\text{B} = \angle\text{C}$ and $\text{AD}\bot\text{BC}.$ The rule by which $\triangle\text{ABD}\cong\triangle\text{ADC}.$
✓
Answer
In $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have
$\angle\text{ABD} = \angle\text{ACD} \ ($given$)$
$\angle\text{BDA} = \angle\text{CDA} \ (90^\circ)$
$AD = AD ($common in both$)$
Hence, $\triangle\text{ABD}\cong\triangle\text{ADC}$ by $\text{AAS}$ criterion.
$\angle\text{ABD} = \angle\text{ACD} \ ($given$)$
$\angle\text{BDA} = \angle\text{CDA} \ (90^\circ)$
$AD = AD ($common in both$)$
Hence, $\triangle\text{ABD}\cong\triangle\text{ADC}$ by $\text{AAS}$ criterion.
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