A Geometry Problem by Mohd.Hamza $II$

Since $BD>AD>CD>ED$ . Therefore, $\frac{\mathrm{area}(ABCD)}{\mathrm{area}(ABCE)} = \frac{\mathrm{area}(\triangle ABC)+\mathrm{area}(\triangle ACD)}{\mathrm{area}(\triangle ABC)+\mathrm{area}(\triangle ACE)}$

Let $A (-1,0); B(1,0); C(x_c,y_c); D(x_d,y_d)$ $AC^2 - BC^2 = (x_c+1)^2 + y_c^2 - (x_c-1)^2 + y_c^2 = 4x_c \in \mathbb{Q}$ $\Rightarrow x_c,x_d \in \mathbb{Q}$ $AC^2+BC^2 = (x_c+1)^2 + y_c^2 +(x_c-1)^2 + y_c^2 = 2y_c^2 \in \mathbb{Q}$ $\Rightarrow y_c^2 , y_d^2 \in \mathbb{Q}$ $CD^2 = (x_c-x_d)^2 + (y_c-y_d)^2 \in \mathbb{Q}$ $\Rightarrow x_c^2+x_d^2-2x_cx_d + y_c^2+y_d^2-2y_cy_d \in \mathbb{Q}$ $\Rightarrow y_cy_d \in \mathbb{Q}$

$\frac{\mathrm{area}(\triangle ACD)}{\mathrm{area}(\triangle ACE)} = |\frac{y_c}{y_d}| = |\frac{y_cy_d}{y_d^2}| \in \mathbb{Q}$ Now, $\frac{\mathrm{area}(\triangle ABC)+\mathrm{area}(\triangle ACD)}{\mathrm{area}(\triangle ABC)+\mathrm{area}(\triangle ACE)} \in \mathbb{Q}$ Since $\mathbb{Z} \subset \mathbb{Q}$ .

Now, lets consider an example:

Let

$\mathrm{area}(\triangle ABC)$ be $1$

$\mathrm{area}(\triangle ACD)$ be $√2$

$\mathrm{area}(\triangle ACE)$ be $2$

Then, $\frac{\mathrm{area}(ABCD)}{\mathrm{area}(ABCE)} = \frac{\mathrm{area}(\triangle ABC)+\mathrm{area}(\triangle ACD)}{\mathrm{area}(\triangle ABC)+\mathrm{area}(\triangle ACE)} = \frac{√2 + 1}{2+1}=\frac{√2 + 1}{3}$ which is irrational .

Therefore Statement 1, 2, 3 all are correct.