Any solution to this question using sine law?

Question

Suppose $ABCD$ is a cyclic quadrilateral and $x, y, z$ are the distances of $A$ from the lines $BD, BC, CD$ respectively. Prove that $\frac{BD}{x} = \frac{BC}{y} + \frac{CD}{x}$

If it can't be proved using law of sines, what are other easy methods?

Answer 1

By Ptolemy's Theorem $\, AC \cdot BD = BC \cdot AD +AB \cdot CD .\, $ Divide both sides by the same value $$ AB \cdot AC \cdot AD \quad \text{to get} \quad \frac{BD}{AB \cdot AD} = \frac{BC}{AB \cdot AC} + \frac{CD}{AC \cdot AD} $$ where the denominators are proportional to $\, x,y,z \,$ respectively. This is because the distance from $\, A \,$ to line $\, BD, \,$ for example, is the length of the perpendicular to that line, and using the law of sines is equal to $\, AB \cdot AD \,$ divided by the diameter of the circle. That is the key result used here.

We prove the result for any triangle $\, ABC. \,$ Drop a perpendicular from $\, A \,$ to the opposite side $\, BC \,$ which splits the triangle into two right triangles which share the perpendicular. By definition of $\, \sin \,$ and the law of sines, the length of the perpendicular is equal to $\, b \sin(C) = c \sin(B) = b \, c / (2R) \,$ where $\, 2R \,$ is the diameter of the circumscribed circle of $\, ABC. \,$