Exercise 15. Let $A B C D$ be a parallelogram with area 1 and $M$ a point on the segment $[B D]$ such that $M D=3 M B$. Let $N$ be the point of intersection of the lines $(A M)$ and $(C B)$. Calculate the area of the triangle $M N D$.


## Solution to Exercise 15

![](https://cdn.mathpix.com/cropped/2024_04_29_bad7460d4ce66061b968g-19.jpg?height=617&width=920&top_left_y=528&top_left_x=568)

In the following, we will denote the area of triangle \(XYZ\) as \(|XYZ|\).

We use the fact that the area of a triangle is equal to half the product of the lengths of its base and its height. Therefore, to compare the areas of different triangles, we can try to find a common base and compare the heights.

First, note that the area of parallelogram \(ABCD\) is equal to twice the area of triangle \(ABD\) because triangles \(ABD\) and \(CDB\) are isometric. Therefore, we have \(|ABD| = \frac{1}{2}\).

We now want to compare the area of triangle \(MND\) with the area of triangle \(ABD\).

Unfortunately, these two triangles do not have a common base. However, triangles \(AND\) and \(ABD\) share the side \(AD\). We draw the heights from vertices \(N\) and \(B\) in triangles \(AND\) and \(ABD\), respectively. Let \(K\) be the foot of the height from vertex \(B\) in triangle \(ABD\) and \(J\) be the foot of the height from vertex \(N\) in triangle \(AND\). We then have

\[
\frac{|AND|}{|ABD|} = \frac{\frac{1}{2} \cdot AD \cdot NJ}{\frac{1}{2} \cdot AD \cdot BK} = \frac{NJ}{BK}
\]

Since the lines \((MI)\) and \((BK)\) are perpendicular to segment \([AD]\), they are parallel. Quadrilateral \(JNBK\) is therefore a parallelogram with two right angles, making it a rectangle. Thus, \(NJ = BK\) and therefore

\[
|AND| = |ABD| = \frac{1}{2}
\]

We apply the same method to compare the areas of triangles \(AND\) and \(AMD\). We introduce \(J\) as the foot of the height from vertex \(N\) and observe that

\[
\frac{|AMD|}{|AND|} = \frac{|AMD|}{|ABD|} = \frac{\frac{1}{2} \cdot AD \cdot MI}{\frac{1}{2} \cdot AD \cdot BK} = \frac{MI}{BK}
\]

The lines \((BK)\) and \((MI)\) are parallel, so by Thales' theorem for triangles \(ABK\) and \(AMI\), we find

\[
\frac{MI}{BK} = \frac{MD}{DB} = \frac{MD}{MB + MD} = \frac{3MB}{MB + 3MB} = \frac{3}{4}
\]

Thus, \(|AMD| = \frac{3}{4} |ABD| = \frac{3}{8}\).

We can then conclude that

\[
|MND| = |AND| - |AMD| = \frac{1}{2} - \frac{3}{8} = \frac{1}{8}
\]

This allows us to conclude that the area of triangle \(MND\) is \(\frac{1}{8}\).