Moving Dot in a Line Segment

Because of the equilateral $\triangle ABC$ ,we can suppose that $AB=BC=AC=x.$ Find the dot $Q$ on $AB$ such that $PQ⊥AB$ and find the dot $R$ on $BC$ such that $PR⊥BC$ .The value $a$ is equal to the sum of the length of $PQ$ and the length of $PR$ .Then $S\triangle ABC=S\triangle PAB+S\triangle PBC=$ $\frac{1}{2}$ $AB\times PQ+$ $\frac{1}{2}$ $BC\times PR=ax.$ And $S\triangle ABC$ is also equal to $\frac{1}{2}$ $AD\times BC=bx.$ Therefore $\frac{1}{2}$ $ax=$ $\frac{1}{2}$ $bx$ , $\boxed{a=b}.$