Maths / Circles / Length of Tangents to a Circle

QUESTION

In the figure given below, $\dpi{120} \fn_jvn \large \angle AOB=90^o\;and\; \angle PAO=\angle QBO.$ Prove that $\dpi{120} \fn_jvn \large \angle AOC=\angle BOC.$

EXPLANATION
Explain TypeExplanation Content
Text

$\\ In\; \Delta OAP \; and \Delta OBQ. \\\\ OP=OQ(radius ) \\\\ \angle OPA=\angle OQB=90^o \\\\ \angle PAO=\angle OBQ(given ) \\\\ \Delta OAP \cong \Delta OBQ(AAS) \\\\ \therefore \: \: OA=OB \\\\ In \; \Delta OCA \; and \; \Delta OCB, \\\\ OA=OB \\\\ OC=OC (common) \\\\ \angle OCA=\angle OCB=90^o \\\\ \therefore \: \: \Delta OCA \cong \Delta OCB \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (by\;RHS\;congruence\;rule)\\\\ \therefore \: \: \angle AOC=\angle BOC$

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