This is a follow on from a question I put here: How to prove $c\sin\theta = a\sin(\theta - B) + b\sin(\theta + A)$
The full question is as follows:$$\\$$ Prove that, for any triangle ABC, with the usual notation and for all values of the angle $\theta$ $$\\c\sin\theta = a\sin(\theta - B) + b\sin(\theta + A)$$
A quadrilateral $ABCD$ is inscribed in a circle. By applying the result above to the triangle $ABC$ and taking $\theta$ for the angle $CAD$, prove that: $$AB\cdot CD + AD\cdot BC = AC\cdot BD$$
I have (with help) managed the first part - the proof. but cannot do the second part. Also, I don’t understand what $\theta$ is. Is it C?
With the sine rule, we have for the inscribed triangles ADC and ADB,
$$CD = 2r\sin\theta, \>\>\>\>\>AD = 2r\sin(B-\theta),\>\>\>\>\>BD=2r\sin(\theta+A)$$
where $r$ is the circumradius. Substitute the above into
$$c\sin\theta = a\sin(\theta - B) + b\sin(\theta + A)$$ to obtain
$$c\cdot CD = -a\cdot AD + b\cdot BD$$
which is just,
$$AB\cdot CD + AD\cdot BC = AC\cdot BD$$