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const:draw AP and DQ perpendicular to BC
Proof:In Triangle AOP and DOQ, Angle AOP=DOQ (VOP) ANGLE APO=DQO=90(BY CONST.)
By AA similarity,AOP similar to DOQ =>AP/DQ=AO/DO ------>(1) ar(ABC)=1/2(BC*AP)
and ar(DBC)=1/2(BC*DQ) ar(ABC)/ar(DBC)=AP/DQ FROM (1),
ar(ABC)/ar(DBC)=AO/DO Hence Proved
Proof:In Triangle AOP and DOQ, Angle AOP=DOQ (VOP) ANGLE APO=DQO=90(BY CONST.)
By AA similarity,AOP similar to DOQ =>AP/DQ=AO/DO ------>(1) ar(ABC)=1/2(BC*AP)
and ar(DBC)=1/2(BC*DQ) ar(ABC)/ar(DBC)=AP/DQ FROM (1),
ar(ABC)/ar(DBC)=AO/DO Hence Proved
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Hii there....
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Dishantgoel:
Mark it as brainlist answer....
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