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Q.

Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at the point O . Then $\frac{OA}{OC}$ = $\frac{OB}{OD}$ (by using similarity criterion for two triangles). Choose whether the given statement is true or false.

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a

True

b

False

answer is A.

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Detailed Solution

Let us first draw the trapezium with the given details.
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Let us consider triangles AOB and COD.
We know that when two lines intersect at a point, the opposite angles at the point of intersection will be equal.
Hence,
∠AOB = ∠COD
We know that if two lines which are parallel are intersected by a transversal then the pair of alternate interior angles are equal. Here, AB ∥ DC and the transversal will be AC. Hence, we get
∠OAB = ∠OCD
Hence, using angle-angle (AA) similarity, we will get
ΔAOB ∼ ΔCOD
We know that when two triangles are similar, the corresponding sides of both the triangles are in proportion to each other. Hence, we will get

\frac{OA}{OC}

=

\frac{OB}{OD}

Hence, proved.
So, option (1) is correct.
Hence, the given statement is true.

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Diagonals AC and BD of a trapezium ABCD with AB ∥ DC intersect each other at the point O . Then OAOC = OBOD (by using similarity criterion for two triangles). Choose whether the given statement is true or false.