Triangle ABC has acute angles with point X internal to ABC. Points P and R are on AB, points T and Q are on BC, and points S and U are on CA so that PQ passes through X and is parallel to AC, RS passes through X and is parallel to BC, and T U passes through X is parallel to BA. I've been asked to show that $\sqrt{|ABC|} = \sqrt{|XPR|} + \sqrt{|XQT|} + \sqrt{|XSU|}.$
I've gotten as far as shown in the image but then defaulted to assuming that the triangle was equilateral to finish. As it was kindly pointed out to me, the the triangle is not specified to be equilateral so I am not sure how to proceed in the case where it is not. Any thoughts on where to go from here?
Let us look for example at the triangle $\triangle{XPR}$. It is similar to $\triangle{CAB}$. Let $a$ be the proportionality constant. Then clearly $\tfrac{A_{\triangle{XPR}}}{A_{\triangle{ABC}}}=a^2$. By denoting the other proportionality constants as $b$ and $c$ for $\triangle{XSU}$ and $\triangle{XTQ}$ respectively, it is clear that the problem asks us to prove $a+b+c=1$.
Now, by definition, $|RX|=a|BC|$, $|XS|=b|BC|$ and $|TQ|=c|BC|$. Note that $BRXT$ is a parallelogram, so $|RX|=|BT|$, meaning that $|BT|=a|BC|$. Similarly, $XSCQ$ is a parallelogram, so $|QC|=|XS|=b|BC|$.
Then, $|BC|=|BT|+|TQ|+|QC|=(a+b+c)|BC|$, so $a+b+c=1$.