Question

the diagonals of a quadrilaeral ABCD intersect each other at point o such that AO/BO=CO/DO

Answer 1

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$\text{\large\underline{\red{correct\:Question:-}}}$

The diagonals of quadilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

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$\text{\large\underline{\orange{Given:-}}}$

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO}$ = $\frac{CO}{DO}$

i.e, $\frac{AO}{CO}$ = $\frac{BO}{DO}$

$\bold{To\: Prove: \:ABCD\: is \:a \:trapezium}$

$\text{\large\underline{\pink{Construction:-}}}$

Draw OE║DC such that E lies on BC

$\text{\large\underline{\green{proof:-}}}$

In ᐃBDC,

$\bold{By \:Basic \:Proportionality\: Theorem,}$

$\frac{BO}{OD}$ = $\frac{BE}{EC}$......$(1)$

But, $\frac{AO}{CO}$ = $\frac{BO}{DO}$.....$(2)$

⛬ $\bold{(1) and (2)}$

$\frac{AO}{CO}$ = $\frac{BE}{EC}$

Hence, By Converse of Basic Proportionality Theorem,

OE║AB

Now Since, AB║OE║DC

⛬ AB║DC

${\fbox{\boxed{\rm{\red{Hence,\:ABCD\: is\: a \:trapezium.}}}}}$

$\text{\large\underline{\pink{Be\: Brainly}}}$

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