If p is any interior point of a triangle

Question

If $p$ is an interior point in a triangle $ABC$, then sin of angle $PAB$ × sin of angle $PBC$ × sin of angle $PCA$ = sin of angle $PBA$ × sin of angle $PCB$ × sin of angle $PAC.$ State and prove the converse of this result .

Source challenge and thrill

Answer 1

Let the triangle be $ABC$

Now let $\angle PAB=\theta_1$ ,$\angle PBC=\theta_2$ ,$\angle PCA=\theta_3$ ,$\angle PBA=\theta_4$ ,$\angle PCB=\theta_5$ ,$\angle PCA=\theta_6$ From $P$ draw perpendiculars on all sides and let $h_1$ be the perpendicular on $AB$, $h_2$ on $AC$ and $h_3$ on $BC.$

Now $$AP=\frac{h_1}{\sin\theta_1}=\frac{h_2}{\sin\theta_6}$$ $$BP=\frac{h_3}{\sin\theta_2}=\frac{h_1}{\sin\theta_4}$$ $$CP=\frac{h_2}{\sin\theta_3}=\frac{h_3}{\sin\theta_5}$$ Multiply all sides and cancel out the perpendiculars and you're left with what you want.

Answer 2

Draw a triangle and you'll find that the product (either left of right) is equal to $$\frac{\text{The product of the distance from P to each side of the triangle}}{\text{The product of the distance from P to each point}}$$

Then use it to prove the converse of the statement.