It is not true. The area of the trapezoid is $a(b+c)/2$. Take the trapezoid $(-2,0), (2,0), (1,1), (-1,1)$. This has area $3$ and diagonal $\sqrt{10}$. If $a=b+c$, the area would be $\frac{a^2}{2}=5$ instead of $3$

It is not true. The area of the trapezoid is $a(b+c)/2$. Take the trapezoid $(-2,0), (2,0), (1,1), (-1,1)$. This has area $3$ and diagonal $\sqrt{10}$. If $a=b+c$, the area would be $\frac{a^2}{2}=5$ instead of $3$

An easy way to see it is to imagine a very wide trapezoid. $b$ and $c$ will be almost vertical and no longer than the height, but the diagonal $a$ is very long.

It is not true. Take the trapezoid $(-2,0), (2,0), (1,1), (-1,1)$. This has area $3$ and diagonal $\sqrt{10}$. If $a=b+c$, the area would be $\frac{a^2}{2}=5$ instead of $3$

It is not true. The area of the trapezoid is $a(b+c)/2$. Take the trapezoid $(-2,0), (2,0), (1,1), (-1,1)$. This has area $3$ and diagonal $\sqrt{10}$. If $a=b+c$, the area would be $\frac{a^2}{2}=5$ instead of $3$