Question

In a trapezium ABCD, AB || CD, AC and BD diagonals are intense
Prove that AO. OD = BO.OC.​

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• In a trap AB//CD
• AC and BD are diagonals

${\sf{\green{\underline{\large{To\: Prove}}}}}$

• AO.OD=BO.OC

${\sf{\pink{\underline{\Large{Explanation}}}}}$

Breaking the given trap in small ∆s

∆BCO

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∆AOD

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Proof

• AO.OD=BO.OC
• we have to proof this
• things so we have to moce to this ∆s
• and similar it

In ∆BCO and ∆AOD

• DAO = OCB (AB//CD and AC is transversal)
• ADO = OBC (AB//CD and BD is transversal)

∆BCO ~ ∆AOD

$\tt\rightarrow\frac{AO}{OC}\frac{OD}{BO}\frac{AB}{CD}$

$\tt\rightarrow\frac{AO}{OC}\frac{OD}{BO}$

=>AO×OD=OC×BO

Hence proved