ABCD is a trapezium with AB parallel to DC. A line parallel to AC intersects AB at X and BC at Y.

Prove that ar(Δ ADX) = ar(Δ ACY).

To prove that the area of triangle ADX is equal to the area of triangle ACY in trapezium ABCD, where AB is parallel to DC, we can use the properties of similar triangles and the concept of parallel lines.

Understanding the Setup

In trapezium ABCD:

• AB is parallel to DC.
• Line XY is drawn parallel to AC, intersecting AB at X and BC at Y.

Properties of Parallel Lines

Since XY is parallel to AC, triangles ADX and ACY are similar by the Basic Proportionality Theorem (also known as Thales' theorem). This means that the ratios of their corresponding sides are equal.

Area of Similar Triangles

The area of similar triangles is proportional to the square of the lengths of their corresponding sides. Therefore, if we denote the lengths of AD and AC as a and b respectively, we have:

Area(Δ ADX) / Area(Δ ACY) = (AD / AC)²

Establishing the Equality

Since AD and AC are parts of the same trapezium and XY is parallel to AC, the segments AX and CY are proportional to the segments AD and AC. Thus, we can conclude:

Area(Δ ADX) = Area(Δ ACY)

Final Thoughts

This proof shows that the areas of triangles ADX and ACY are equal due to the properties of parallel lines and similar triangles within trapezium ABCD.