The diagonals of quadrilateral ABCD intersect at O- prove that ar ACB ar ACD - BO DO

ABCD is a quadrilateral intersect at O .Let BO⊥ AC and DO⊥ AC ∴ #160; ar( ACB)=12× BO× AC #160; #160;— ( 1 ) ∴ # ...

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Standard IX

Mathematics

Relationship between Unequal Sides of Triangle and the Angles Opposite to It.

The diagonals...

Question

The diagonals of quadrilateral ABCD intersect at O, prove that $\frac{a r (△ A C B)}{a r (△ A C D)} = \frac{B O}{D O}$

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Solution

$A B C D$ is a quadrilateral intersect at $O$ .

Let $B O ⊥ A C$ and $D O ⊥ A C$
$∴$ $a r (△ A C B) = \frac{1}{2} \times B O \times A C$ --- ( 1 )

$∴$ $a r (△ A C D) = \frac{1}{2} \times D O \times A C$ --- ( 2 )

Now, dividing ( 1 ) by ( 2 ),

$\frac{a r (△ A C B)}{a r (△ A C D)} = \frac{\frac{1}{2} \times B O \times A C}{\frac{1}{2} \times D O \times A C}$

$∴$ $\frac{a r (△ A C B)}{a r (△ A C D)} = \frac{B O}{D O}$

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The diagonals of quadrilateral ABCD intersect at O, prove that ar(△ACB) ar(△ACD) =BODO

ABCD is a quadrilateral intersect at O.Let BO⊥AC and DO⊥AC∴ ar(△ACB)=12×BO×AC --- ( 1 )∴ ar(△ACD)=12×DO×AC --- ( 2 )Now, dividing ( 1 ) by ( 2 ),ar(△ACB)ar(△ACD)=12×BO×AC12×DO×AC∴ ar(△ACB)ar(△ACD)=BODO

The diagonals of quadrilateral ABCD intersect at O, prove that $\frac{a r (△ A C B)}{a r (△ A C D)} = \frac{B O}{D O}$