proof for trapezoid within triangle conundrum

Question

I've recently been working on some geometry problems involving non-isosceles trapezoids and stumbled upon an interesting discovery which seems to hold true for all cases I've tried, but I don't have the expertise to construct a proof. Here's the thing:

1. take any random triangle ABC
2. choose two points Q on edge AB and P on edge AC such that QP || BC. We now have constructed a trapezoid QBCP

3. find K the intersection point of the trapezoid's diagonals.

4. The line AK will perfectly bisect QP and BC in the middle. This is what requires proof.

I am very interested if somebody can construct a proof for this. Thanks

illustration

Answer 1

Let $AK\cap QP=\{M\}$, $AK\cap BC=\{N\}$, $QM=a$, $MP=b$, $BN=c$ and $NC=d$.

Thus, by similarity of triangles we obtain: $$\frac{a}{c}=\frac{AM}{AN}=\frac{b}{d}$$ and $$\frac{a}{d}=\frac{MK}{KN}=\frac{b}{c},$$ which gives $a=b$ and $c=d$.