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Inequality for sides of a triangles
I recently encountered this geometric problem (I could not find a better title).
Let $ABC$ a triangle and $P$ a point on the segment $AB$. Let $Q$ be
the intersection of the line $CP$ with the circumscribed circle of the
triangle different from $C$. Show the inequality
$$\frac{\overline{PQ}}{\overline{CQ}} \le \Big(\frac
{\overline{AB}}{\overline{AC}+\overline{CB}}\Big)^2 .$$ Further,
show that equality holds if and only if $CP$ is the angle bisector of
the angle $\angle ACB$.
As in most problems of this kind, finding the solution is a matter drawing the right lines. I already come up with a solution. I wanted to share the problem as there are probably many different ways to solve it.
To be clear: I don't need help with it, I just thought maybe there are some people here who like to do this kind of problems and want to try themselves on it.
First, let $CPQ$ bisect $\angle ACB$, and join $AQ$, $BQ$. Then $AQ=BQ$, and by Ptolemy's Theorem $$AB\

First, let $CPQ$ bisect $\angle ACB$, and join $AQ$, $BQ$. Then $AQ=BQ$, and by Ptolemy's Theorem $$AB\times CQ=AQ\times(CA+CB)$$Dividing both sides by $AC\times AQ$, inverting and squaring$$\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$Thus we must show that $$\frac{PQ}{CQ}=\frac{AQ^2}{CQ^2}=\frac{AB^2}{(AC+CB)^2}$$or$$PQ\times CQ=AQ^2$$But since $\angle ACB$ is bisected, $\angle ACQ=\angle QAB$, making triangles $ACQ$ and $QAB$ similar.
Hence$$\frac{PQ}{AQ}=\frac{AQ}{CQ}$$or$$PQ\times CQ=AQ^2$$
Thus the ratios are equal when $\angle ACB$ is bisected.
Now let $R$ be any other point on $AB$, and draw $CR$ thru to $S$ on the circumference, and join $QS$.
Then since arcs $AQ$, $BQ$ are equal, the tangent at $Q$ is parallel to $AB$, and therefore chord $QS$ is inclined toward $PR$, making$$\frac{SR}{RC}<\frac{QP}{PC}$$Hence$$\frac{SR}{SR+RC}<\frac{QP}{QP+PC}$$or$$\frac{SR}{SC}<\frac{QP}{QC}$$
And since $\frac{AB^2}{(AC+CB)^2}$ is fixed$$\frac{SR}{SC}<\frac{AB^2}{(AC+CB)^2}$$
20171203 05:37:36