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Question

# In the given figure, PA ⊥ AB, QB ⊥ AB and PA = QB. Prove that ∆OAP ≅ ∆OBQ. Is OA = OB? Figure

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Solution

## $\mathrm{Given}:\phantom{\rule{0ex}{0ex}}PA\perp AB\phantom{\rule{0ex}{0ex}}QB\perp AB\phantom{\rule{0ex}{0ex}}PA=QB\phantom{\rule{0ex}{0ex}}\mathrm{To}\mathrm{prove}:△OAP\cong △OBQ\phantom{\rule{0ex}{0ex}}\mathrm{Find}\mathrm{whether}OA=OB.\phantom{\rule{0ex}{0ex}}\mathrm{Proof}:\phantom{\rule{0ex}{0ex}}\mathrm{In}△OAP\mathrm{and}△OBQ:\phantom{\rule{0ex}{0ex}}\angle POA=\angle QOB\left(v\mathrm{ertically}\mathrm{opposite}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle OAP=\angle OBQ\left(90°\mathrm{each}\right)\phantom{\rule{0ex}{0ex}}PA=QB\left(g\mathrm{iven}\right)\phantom{\rule{0ex}{0ex}}ByAAS\mathrm{congruence}\mathrm{property}:\phantom{\rule{0ex}{0ex}}△OAP\cong △OBQ\phantom{\rule{0ex}{0ex}}=>OA=OB\left(\mathrm{corresponding}\mathrm{parts}\mathrm{of}\mathrm{the}\mathrm{congruent}\mathrm{triangles}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

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