Question

State "Basic Proportionality Theorem" and use it to prove the following:

In a quadrilateral ABCD, diagonals AC and BD intersect each other at O such that $\frac{A O}{BO} = \frac{CO}{D O}$ as shown in the given figure. Prove that ABCD is a trapezium.

Approach

State the Basic Proportionality Theorem (BPT) and use it to prove that in quadrilateral ABCD, if diagonals AC and BD intersect at O such that $\frac{A O}{BO} = \frac{CO}{D O}$ , then ABCD is a trapezium. BPT: If a line is drawn parallel to one side of a triangle intersecting the other two sides, the other two sides are divided in the same ratio. Construct OE parallel to AB. In triangle DAB, by BPT, $\frac{D E}{A E} = \frac{D O}{BO}$ . Given $\frac{A O}{BO} = \frac{CO}{D O}$ , so $\frac{D O}{BO} = \frac{CO}{A O}$ . Thus $\frac{D E}{A E} = \frac{CO}{A O}$ . In triangle ADC, by converse of BPT, OE is parallel to CD. Since OE is parallel to both AB and CD, AB is parallel to CD, making ABCD a trapezium.

Step-by-step working

1
State BPT: If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, the other two sides are divided in the same ratio.
1 mark
2
Given: Quadrilateral ABCD with diagonals intersecting at O, $\frac{A O}{BO} = \frac{CO}{D O}$ . To Prove: AB $∥$ CD. Construction: Draw OE $∥$ AB.
1 mark
3
In $△ D A B$ , OE $∥$ AB, so by BPT: $\frac{D E}{A E} = \frac{D O}{BO}$ .
1 mark
4
Given $\frac{A O}{BO} = \frac{CO}{D O}$ , so $\frac{D O}{BO} = \frac{CO}{A O}$ . Hence $\frac{D E}{A E} = \frac{CO}{A O}$ .
1 mark
5
In $△ A D C$ , $\frac{D E}{A E} = \frac{CO}{A O}$ , so by converse of BPT, OE $∥$ CD. Since OE $∥$ AB and OE $∥$ CD, AB $∥$ CD. Thus ABCD is a trapezium.
1 mark

Concepts used

basic proportionality theoremthales theoremprovetrapeziumparallel lines

Approach

Step-by-step working

Concepts used

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