Question
In fig. (i) $\text{PL}\bot\text{OA}$ and $\text{PM}\bot\text{OB}$ such that $\text{PL}=\text{PM}$. Is $\triangle\text{PLO}\cong\triangle\text{PMO}?$ Give reason in support of your answer. 
✓
Answer
In fig.$\text{PL}\bot\text{OA}$ and $\text{PM}\bot\text{OB}$ and $\text{PL}=\text{PM}$
Now in right $\triangle\text{PLO}$ and $\triangle\text{PMO}$,
Side $\text{PL}=\text{PM}$ (given)
Hypotenuse $\text{OP}=\text{OP}$ (common)
$\triangle\text{PLO}\cong\triangle\text{PMO} (RHS$ condition$)$
Yes $\triangle\text{PLO}\cong\triangle\text{PMO}$
Hence proved.
Now in right $\triangle\text{PLO}$ and $\triangle\text{PMO}$,
Side $\text{PL}=\text{PM}$ (given)
Hypotenuse $\text{OP}=\text{OP}$ (common)
$\triangle\text{PLO}\cong\triangle\text{PMO} (RHS$ condition$)$
Yes $\triangle\text{PLO}\cong\triangle\text{PMO}$
Hence proved.
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