Prove that $|AM|=|BM|$, if ...

Question

Let $M$ be the point of intersection of the diagonal sides of a trapezoid.

Let $l$ be the line through $M$ that is parallel to the bases of the trapezoid.

Let $A$ and $B$ be the points in which the diagonals of the trapezoid cut $l$.

I have to prove that $AM=BM$.

Can you give me a step by step solution?

Answer 1

In the diagram, side $\overline{PQ}$ is parallel to side $\overline{RS}$.

One theorem in geometry is that if a line parallel to a side of a triangle intersects the other two sides then the new triangle is similar to the original triangle. The similar triangles create a variety of equal proportions. Using that theorem repeatedly we get

$$\begin{align} \frac{SR}{AM} & =\frac{QR}{QM} \qquad\text{by line SR through triangle QAM}\\ & =\frac{PS}{PM}\qquad\text{by line SR through triangle MPQ}\\ & =\frac{SR}{BM}\qquad\text{by line SR through triangle PBM} \end{align}$$

Equating the first and last expressions, we get

$$AM=BM$$

Answer 2

Since any affine map preserves the ratios of segments on the same line we can assume without loss of generality that the trapezoid is an isosceles trapezoid. Then the claim is trivial.