MCQ
- AAAS
- BSAS
- CSSS
- DRHS
✓
Answer
- AAS
Solution:
In $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have
$\angle\text{ABD} = \angle\text{ACD} $ (given)
$\angle\text{BDA} = \angle\text{CDA} $ (90o)
AD = AD (common in both)
Hence, $\triangle\text{ABD}\cong\triangle\text{ADC}$ by AAS criterion.
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