Proving: "a quadrilateral can be inscribed in a circle iff the product of its diagonals equals the sum of the products of its opposite sides"
The book I'm following uses this diagram and goes on to show:
(1) We can express:
$$AB \cdot CD + AD \cdot BC = AC \cdot BD$$
(2) Using:
$$\sin\alpha \sin\beta + \sin\gamma\sin\delta = \sin(\beta+\gamma)\sin(\alpha+\gamma)$$
Further, the book goes on derive "Ptolemy's Identity" from this:
If $\alpha + \beta + \gamma + \delta = \pi$, then $\sin\alpha \sin\beta + \sin\gamma\sin\delta = \sin(\beta+\gamma)\sin(\alpha+\gamma)$
This statement is equivalent to the part of Ptolemy's theorem that says if a quadrilateral is inscribed in a circle, then the product of the diagonals equals the sum of the products of the opposite sides.
I somehow can't follow the proof completely, because:
- I don't understand what rewriting the equation from (1) to (2) actually shows. It just shows that I can express the equality using sine of the angles, but how does it make the theorem true?
- In the identity described, how is the sum of the angles equal to $\pi$ connected to the theorem and why are these statements equivalent, as claimed above?
Using sine rule we have $${AB\over \sin \alpha} = {AC\over \sin (\gamma+\delta)} = {AD\over \sin \gamma}= {BC\over \sin \delta}={BD\over \sin (\alpha+\gamma)} = {CD\over \sin \beta}=2r $$
where $2r$ is a diameter of a given circle.
Now expres the sides and plug it in a $(1)$ and you will get $(2)$. Now you have to prove that $(2)$ is actualy true following the rules of trigonometry.