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Answer 1

$\large\underline{\sf{Given \:Question - }}$

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

$\large\underline{\sf{Solution-}}$

Given that,

The diagonals of a quadrilateral ABCD intersect each other at the point O such that

$\rm :\longmapsto\:\dfrac{AO}{BO} = \dfrac{CO}{DO}$

Now, Consider,

$\red{\rm :\longmapsto\:\: \triangle \: AOB \: and \:\triangle \: COD}$

$\rm :\longmapsto\:\dfrac{AO}{BO} = \dfrac{CO}{DO} \: \: \: \: \: \: \: \: \: \{ \: given \: \}$

$\rm :\longmapsto\:\angle \: AOB \: = \: \angle \: COD \: \: \: \: \{vertically\:opposite \: angles \}$

$\red{\rm :\longmapsto\:\: \triangle \: AOB \: \sim \:\triangle \: COD \: \: \: \: (SAS \: Similarity)}$

$\bf\implies \:\angle \: BAO \: = \: \angle \: DCO \: \: \: \: \: (CPST)$

But these are Alternate interior angles

Thus,

$\bf\implies \:AB \: \parallel \: CD$

$\bf\implies \:ABCD \: is \: a \: trapezium$

Hence, Proved

Additional Information

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.

4. Basic Proportionality Theorem,

If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

Answer 2

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

_____________________________________

Given

• Quadrilateral ABCD where AC and BD intersects each other at O such that,

$\sf\: \frac{ AO}{BO }= \frac{CO}{DO}$

To prove

• ABCD is a trapezium

Construction

• From the point O, draw a line EO touching AD at E, in such a way that, EO || DC || AB

Solution

In ΔDAB, EO || AB

By using Basic Proportionality Theorem

$\sf \: \frac{ DE}{EA} = \frac{DO}{OB }…(i)$

Also, given,

$\sf\: \frac{ AO}{BO }= \frac{CO}{DO}$

$\tt ⇒ \frac{AO}{CO} = \frac{ BO}{DO } \\ \\ \tt ⇒ \frac{ CO}{AO} = \frac{DO}{BO} \\ \\ \tt ⇒ \frac{ DO}{OB} = \frac{ CO}{AO} ...(ii)$

From equation (i) and (ii)

we obtain

$\sf \: \frac{ DE}{EA} = \frac{CO}{AO}$

Therefore, By applying converse of Basic Proportionality Theorem,

EO || DC

Also EO || AB ⇒ AB || DC.

So, quadrilateral ABCD is a trapezium with AB || CD.

Hence Proved

$\bold \red{Celestial} \bold{Centrix}$