Proof of Area Proportionality in Trapezium Diagonals

Introduction to Trapezium Areas and Parallelism

In the realm of geometric proofs, one fascinating scenario concerns the relationship between areas formed by diagonals in a trapezium. Particularly interesting is the proof that in a trapezium (ABCD) where (AB parallel CD) and diagonals (AC) and (BD) intersect at point (O), the following relationship holds:

Area(AOB) × Area(DOC) Area(AOD) × Area(BOC)

This article delves into the detailed, step-by-step proof of this relationship, highlighting the essential geometric principles and properties utilized.

Step-by-Step Proof

Identify the Areas

Let us denote the areas of the triangles formed by the diagonals as follows:

(S_1 text{Area}(AOB)) (S_2 text{Area}(BOC)) (S_3 text{Area}(COD)) (S_4 text{Area}(DOA))

Property of Parallel Lines

Given that (AB parallel CD), the triangles (AOB) and (COD) are similar, as are the triangles (BOC) and (DOA). This similarity is due to the equal angles formed at point (O):

(angle AOB angle COD) (angle BOC angle DOA)

Area Ratios

The similarity of these triangles allows us to express their areas in terms of their bases and heights. Considering the height from point (O) to lines (AB) and (CD), which are the same for the respective triangles, we can write:

(frac{S_1}{S_3} frac{AB}{CD}) (frac{S_2}{S_4} frac{AB}{CD})

Relate Areas Using Ratios

From these similarity ratios, we can express the areas:

(S_1 S_3 cdot frac{AB}{CD}) (S_2 S_4 cdot frac{AB}{CD})

Substituting these into the equation we want to prove:

(S_1 times S_3 S_2 times S_4)

Substituting the Areas

Substitute the expressions for (S_1) and (S_2):

(S_3 cdot frac{AB}{CD} times S_3 S_4 cdot frac{AB}{CD} times S_4)

This simplifies to:

(frac{AB}{CD} cdot S_3^2 frac{AB}{CD} cdot S_4^2)

Since (frac{AB}{CD}) is common and non-zero, it can be canceled out:

(S_1 times S_3 S_2 times S_4)

This confirms the relationship we set out to prove:

(text{Area}(AOB) times text{Area}(DOC) text{Area}(AOD) times text{Area}(BOC))

Further Analysis

For a deeper understanding, let us introduce the bases (AB b_1) and (CD b_2), and the total height (h). The heights from point (O) to the lines (AB) and (CD) are in the ratio of their bases:

(frac{h_1}{h_2} frac{AB}{CD})

Expressing the heights in terms of the bases:

(h_1 h cdot frac{b_1}{b_1 b_2})

(h_2 h cdot frac{b_2}{b_1 b_2})

Using these relations, we can calculate the areas:

(text{Area}(AOB) frac{1}{2} h_1 b_1) (text{Area}(BOC) text{Area}(ABC) - text{Area}(AOB) frac{1}{2} h b_1 - frac{1}{2} h_1 b_1 frac{1}{2} b_1 h - frac{1}{2} frac{h b_1 b_2}{b_1 b_2} frac{1}{2} b_1 h frac{b_2}{b_1 b_2}) (text{Area}(COD) frac{1}{2} h_2 b_2) (text{Area}(AOD) text{Area}(ABD) - text{Area}(AOB) frac{1}{2} h b_1 - frac{1}{2} h_1 b_1 frac{1}{2} b_1 h frac{b_2}{b_1 b_2})

Calculating the product of the areas of (AOB) and (DOC):

(text{Area}(AOB) times text{Area}(DOC) left(frac{1}{2} h_1 b_1right) left(frac{1}{2} h_2 b_2right) frac{1}{4} h_1 h_2 b_1 b_2 frac{1}{2} b_1 b_2 h/left(b_1 b_2right)^2)

The product of the areas of (AOD) and (BOC):

(text{Area}(AOD) times text{Area}(BOC) left(frac{1}{2} b_1 h frac{b_2}{b_1 b_2}right) left(frac{1}{2} b_1 h frac{b_2}{b_1 b_2}right) frac{1}{4} b_1 b_2 h/left(b_1 b_2right)^2)

Hence, the equality holds true: