In the cyclic quadrilateral $ABCD$, $AB:BC:CD:DA=1:9:9:8$, $AC$ intersects $BD$ at $P$, what is $S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCD}:S_{\triangle PDA}$?
I have no idea how to start this question; how do I get the areas of the triangles when only given side lengths? Please help me out.
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Note,
$$\frac{PB}{PD}=\frac{S_{\triangle ABC}}{S_{\triangle ACD}} = \frac{\frac12\cdot AB\cdot BC\cdot \sin B}{\frac12\cdot AD\cdot DC\cdot \sin D} = \frac{ AB\cdot BC}{ AD\cdot DC}=\frac{1}{8}$$
Similarly,
$$\frac{PA}{PC}=\frac{ AB\cdot AD}{ BC\cdot DC}=\frac{8}{81}$$
Recognize
$$\frac{S_{\triangle PAB}}{S_{\triangle PAD}} =\frac{S_{\triangle PBC}}{S_{\triangle PCD}} =\frac{PB}{PD} = \frac18,\>\>\>\>\> \frac{S_{\triangle PAB}}{S_{\triangle PBC}}=\frac{PA}{PC} = \frac8{81} $$
to obtain,
$$S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCD}:S_{\triangle PDA} =8:81:648:64$$
Consider $\triangle PAB$ and $\triangle PDC$. These pairs of inscribed angles are equal:
$$\begin{align*} \angle CAB &=\angle BDC\\ \angle ABD &= \angle DCA \end{align*}$$
So $\triangle PAB \sim\triangle PDC$. The ratio of their areas are square of the ratio of their corresponding sides:
$$\frac{S_{\triangle PAB}}{S_{\triangle PDC}} = \left(\frac{AB}{DC}\right)^2 = \frac{1}{81}$$
Similar, for $\triangle PBC$ and $\triangle PAD$, consider these pairs of inscribed angles,
$$\begin{align*} \angle DBC &= \angle CAD\\ \angle BCA &= \angle ADB\\ \triangle PBC &\sim \triangle PAD\\ \frac{S_{\triangle PBC}}{S_{\triangle PAD}} &= \left(\frac{BC}{AD}\right)^2\\ & = \frac{81}{64} \end{align*}$$
Between $\triangle PAB$ and $\triangle PAD$, these inscribed angles are equal because $BC = CD$:
$$\angle CAB = \angle CAD$$
Then the ratio of their areas is
$$\begin{align*} \frac{S_{\triangle PAB}}{S_{\triangle PAD}} &= \frac{\frac12 PA\cdot AB\sin\angle PAB}{\frac12PA\cdot AD\sin PAD}\\ &= \frac{AB}{AD}\\ &= \frac 18\\ \end{align*}$$
$$S_{\triangle PAB}:S_{\triangle PBC}:S_{\triangle PCD}:S_{\triangle PDA} = 8:81:648:64 $$