geometry
Let $ABCD$ be a cyclic quadrilateral with side lengths $AB=p,BC=q,CD=r,DA=s$.Show that $\dfrac{AC}{BD}=\dfrac{ps+qr}{pq+rs}$.
My work:
I have found out that this follows from Ptolemy's Second Theorem but cannot prove it.Please help! With or without Ptolemy is fine, I do not have any restriction.
Let $x\approx 4.45171859943238$ be the root of the function $$f(t)=\frac{2}{t-\sqrt 3} - \frac{4\sqrt{t^2+1}}{t^2+5}.$$ Let $ABC$ be an equilateral triangle with sides of length $2$. Let $D$ be a point on the same side of line $AC$ as $B$ such that $AD=CD=\sqrt{x^2+1}$. Then the distance of $D$ from $AC$ is $x$, hence $BD=x-\sqrt 3$. We have $$\frac{AC}{BD} = \frac{2}{x-\sqrt 3} = \frac{4\sqrt{x^2+1}}{x^2+5} = \frac{AB \cdot AD + CB \cdot CD}{BA \cdot BC + DA \cdot DC},$$ meaning that $ABCD$ is a counterexample to the statement you want to prove.
Best Answer
As you've said this follows immidiatelly from the Second Ptolemy Therem. Here's the proof for it.
Note that in every triangle we have:
$$bc = 2Rh_{a}$$
This follows from the formula for area of triangle: $P = \frac{abc}{4R}$
Note: Vertex C and vertex D need to switch places.
So from $\triangle ABD$ and $\triangle BCD$ in the picture we have:
$$ad = 2Rh_1 \quad \quad \text { and } \quad \quad {bc = 2Rh_2}$$
Adding this we have:
$$ad + bc = 2Rh_1 + 2Rh_2$$
From the right trinagles, where $AK$ and $KD$ are hypothenyses we have:
$$h_1 = AK \cdot \sin w \quad \quad \text { and } \quad \quad {h_2 = KC \cdot \sin w}$$
Substitunting we have:
$$ad + bc = 2R \sin w (AK + KC) = 2R \cdot AC \sin w$$
Simularly we have:
$$ab + cd = 2R \cdot BD \sin (\pi - w)$$
But $\sin w = \sin (\pi - w)$, so we have:
$$\frac{ad + bc}{ab + cd} = \frac{2R \cdot AC \sin w}{2R \cdot BD \sin (\pi - w)} = \frac{AC}{BD}$$