MCQ
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}.
$
$
- ✓$AAS$
- B$SAS$
- C$SSS$
- D$RHS$
✓
Answer
Correct option: A.
$AAS$
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 $AAS$ 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 $AAS$ criterion.
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