The altitudes BQ and CP of D ABC meet at O.
Prove that:
(i) CQ ∙ OP = BP ∙ OQ
(ii) D POQ ~ D BOC
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Given : The altitudes BQ and CP of Δ ABC meet at O.
To Find : Prove that:
(i) CQ ∙ OP = BP ∙ OQ
(ii) Δ POQ ~ Δ BOC
Solution:
Compare ΔOQC and Δ OPB
∠OQC = ∠OPB = 90°
∠QOC = ∠POB ( vertically opposite angle)
=> ΔOQC ≈ Δ OPB (AA)
=> OQ/OP = CQ/BP
=> OQ . BP = OP . CQ
=> CQ . OP = BP. OQ
QED
Hence Proved
ΔOQC ≈ Δ OPB
=> OQ/OP = OC/OB
=> OQ/OC = OP/PB
Compare Δ POQ and Δ BOC
OQ/OC = OP/PB
∠POQ = ∠BOC
=> Δ POQ ≈ Δ BOC ( SAS )
QED
Hence proved
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goggle it boy sorry about u
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