Question

The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{A O}{OC} = \frac{BO}{O D}$ . Show that quadrilateral ABCD is a trapezium.

Approach

In quadrilateral ABCD with diagonals intersecting at O such that $\frac{A O}{OC} = \frac{BO}{O D}$ , show ABCD is a trapezium. Since triangles AOB and COD share angle at O and have proportional sides, they are similar, so $∠ O A B = ∠ OC D$ . These are alternate angles, implying DC is parallel to AB, making ABCD a trapezium.

Step-by-step working

1
Draw correct figure showing quadrilateral ABCD with diagonals intersecting at O
0.5 marks
2
$\frac{A O}{OC} = \frac{BO}{O D}$ and $∠ A OB = ∠ CO D$ (vertically opposite) $\Rightarrow △ A OB \sim △ CO D$
1 mark
3
$∠ O A B = ∠ OC D$ (corresponding angles in similar triangles). These are alternate angles, so DC $∥$ AB. Therefore, ABCD is a trapezium
0.5 marks

Concepts used

trianglessimilar trianglestrapeziumratio of sidesprove

Approach

Step-by-step working

Concepts used

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