Question

the diagonals of a quadrilateral ABCD intersect each other at the point O, such that AO/BO=CO/DO. show that ABCD os a trapezium.

Answer 1

FIGURE 1

Step-by-step explanation:$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$$\: \: \: { \orange{given}} \\{ \pink{ \boxed{ \green{ \frac{AO}{BO} = \frac{CO}{DO} }}}} \\ \\ { \blue{to \: prove}} \\ { \purple{ \boxed{ \red{AB || CD}}}}$☆ PROOF:$In \triangle ABD \\ OE || AB \: \: \: (construction) \\ \frac{AE}{DE} = \frac{BO}{DO} \: \: \: \: \: \: (by \: thales \: theoram) - - - - - (1)</p><p>\\ again \\ \frac{AO}{BO} = \frac{CO}{DO} \: \: \: \: \: \: (given) \\ \to \frac{AO}{CO} = \frac{BO}{DO} - - - - - (2)$☆ FROM (1) AND (2)$\frac{AE}{DE} = \frac{AO}{CO} \\ \therefore EO || DC \: \: \: \: (by \: opposite \: of \: thales \: theoram) \\ \\ so \: AB || CD \\ { \pink{ \boxed{ \green{PROVED}}}}$

Answer 2

Given :

The diagonals of a quadrilateral ABCD intersect each other at the point O, such that AO/BO=CO/DO .

To show :

Show that ABCD os a trapezium.

Solution :

Here , we will be basically proving this , by using the mid point theorem .

According to the midpoint Theorem :

If a line is drawn through the mid-point of a side of a triangle parallel to the second side, it will bisect the third side.

But , as we can observe , mid point theorem is valid only for triangles .

So , we will first join the diagonals of the quadrilateral ABCD to create a triangle .

The intersection point of the diagonals is O .

See the attached figure for more details ..

Now, after this the proof will be more or less general ...

AE / DE = BO / DO = AO / CO

Hence , EF // AB // DC // EO

This implies one pair of opposite sides are parallel .

Hence , ABCD becomes a trapezium .

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