A geometry problem by Mohd. Hamza

The proof is similar to this: $A (-1,0); B(1,0); C(x_c,y_c); D(x_d,y_d)$ $AC^2 - BC^2 = (x_c+1)^2 + y_c^2 - (x_c-1)^2 + y_c^2 = 4x_c \in \mathbb{Q}$ $\Rightarrow x_c,x_d \in \mathbb{Q}$ $AC^2+BC^2 = (x_c+1)^2 + y_c^2 +(x_c-1)^2 + y_c^2 = 2y_c^2 \in \mathbb{Q}$ $\Rightarrow y_c^2 , y_d^2 \in \mathbb{Q}$ $CD^2 = (x_c-x_d)^2 + (y_c-y_d)^2 \in \mathbb{Q}$ $\Rightarrow x_c^2+x_d^2-2x_cx_d + y_c^2+y_d^2-2y_cy_d \in \mathbb{Q}$ $\Rightarrow y_cy_d \in \mathbb{Q}$ $\frac{\mathrm{area}(\triangle ACD)}{\mathrm{area}(\triangle ACE)} = |\frac{y_c}{y_d}| = |\frac{y_cy_d}{y_d^2}| \in \mathbb{Q}$ Proving each ratio as rational will result in their summation being rational.

QED

This isn’t an explanation, but I saw a very very similar problem that was 6A on a Putnam that asks to prove that the ratio of areas of two triangles under these same circumstances is rational. That problem drove me a bit crazy and I still haven’t figured it out, I should work on it though.