Answer

  1. RHS
    Solution:
    In $\triangle\text{ABD}$ and $\triangle\text{ACD},$ we have
    $\angle\text{ADB} = \angle\text{ADC}$ (Right angles)
    AB = AC (Given and hypotenuses)
    AD = AD (common in both)
    Therefore, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by RHS.

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