State "Basic Proportionality Theorem" and use it to prove the follo...

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 $\dfrac{AO}{BO}=\dfrac{CO}{DO}$. Prove that $ABCD$ is a trapezium.

Accepted Answer

Step 1: Basic Proportionality Theorem (BPT / Thales Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then the other two sides are divided in the same ratio. Step 2: Given: Quadrilateral $ABCD$ with diagonals $AC$ and $BD$ meeting at $O$, and $$\frac{AO}{BO}=\frac{CO}{DO}$$ Step 3: To prove: $ABCD$ is a trapezium, i.e. $AB \parallel DC$. Step 4: Construction: Through $O$, draw $OE \parallel AB$ meeting $AD$ at $E$. Step 5: In $ riangle DAB$, since $OE \parallel AB$, by BPT $$\frac{DE}{EA}=\frac{DO}{OB} \quad \cdots (i)$$ Step 6: The given condition $\dfrac{AO}{BO}=\dfrac{CO}{DO}$ can be rewritten as $$\frac{AO}{CO}=\frac{BO}{DO} \;\Rightarrow\; \frac{CO}{AO}=\frac{DO}{BO} \quad \cdots (ii)$$ Step 7: From (i) and (ii), $$\frac{DE}{EA}=\frac{CO}{AO}$$ Step 8: In $ riangle ADC$, the line $EO$ divides the sides $AD$ and $AC$ in the same ratio, so by the converse of BPT $$EO \parallel DC$$ Step 9: But by construction $EO \parallel AB$. Two lines parallel to the same line are parallel to each other. $$AB \parallel DC$$ Step 10: A quadrilateral with one pair of opposite sides parallel is a trapezium. Hence $ABCD$ is a trapezium. Proved.

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