Show that $BP+BQ=2PQ $ Let consider a circle of diameter $CA $ and $B\in CA $ such that $A\in [CB] $ and $AB=\frac {CA}{2} $.
If $M \in [CA] $ such that $AM=\frac {CA}{3} $ and $P, Q $ on circle such that $P, M, Q $ collinear.
Show that $BP+BQ=2PQ $
My idea. I notice that $M $ is the middle of $[CB] $. I take the middle of $[CP] $ and $[CQ] $. Now I am stuck.
 A: The problem (with $AM=\frac{CA}{3}$) cannot be correct.  Choose $P$ and $Q$ so that $PQ\perp CA$ at $M$.  If $R$ is the radius of the circle, then we have $$OM=\frac{R}{3}\ \Rightarrow\ PM=\sqrt{OP^2-OM^2}=\sqrt{R^2-\left(\frac{R}{3}\right)^2}=\frac{2\sqrt{2}}{3}R.$$
This gives $PQ=2\ PM=\frac{4\sqrt{2}}{3}R$.  On the other hand,
$$BM=AB+AM=\frac{CA}{2}+\frac{CA}{3}=\frac{2R}{2}+\frac{2R}{3}=\frac{5}{3}R,$$
so
$$BP=\sqrt{BM^2+PM^2}=\sqrt{\left(\frac{5}{3}R\right)^2+\left(\frac{2\sqrt{2}}{3}R\right)^2}=\frac{\sqrt{33}}{3}R.$$
By symmetry, $BQ=BP=\frac{\sqrt{33}}{3}R$.  Clearly,
$$BP+BQ=\frac{2\sqrt{33}}{3}R\neq \frac{8\sqrt{2}}{3}R=2\ PQ.$$

What may be the correct condition is that $AM=\frac{CA}{4}$.  In this case, $A$ and $C$ harmonically divide $B$ and $M$.  Thus, if $\Gamma$ is the circle with diameter $CA$ and $X\in \Gamma$, then $AX$ and $CX$ are the internal and external angular bisectors of $\angle BXM$, respectively.  This proves that $A$ is the incenter of the triangle $BPQ$.  This means $AB$ internally bisects $\angle BPQ$.

Let $Q'$ be the reflection of $Q$ about the line $CA$.  Then, $Q'$ is on the line $BP$ and also on the circle $\Gamma$.  Therefore,
$$BP\cdot BQ=BP\cdot BQ'=BA\cdot BC= R\cdot 3R=3R^2,$$
if $R$ denotes the radius of $\Gamma$.  This shows that 
$$\frac{BM^2}{BP\cdot BQ}=\frac{\left(\frac{3}{2}R\right)^2}{3R^2}=\frac{3}{4}.$$
It is well-known (via Stewart's theorem and the angle bisector theorem, for example) that
$$BM^2=BP\cdot BQ\left(1-\frac{PQ^2}{(BP+BQ)^2}\right).$$
That is
$$1-\frac{PQ^2}{(BP+BQ)^2}=\frac{3}{4},$$
so $BP+BQ=2\ PQ$, as desired.
Or as Lozenges remarked, by the angle bisector theorem, $$\frac{BP}{PM}=\frac{BA}{AM}=2$$ and
$$\frac{BQ}{QM}=\frac{BA}{AM}=2.$$
So, $BP=2\ PM$ and $BQ=2\ QM$, making $$BP+BQ=2\ PM+2\ QM=2\ PQ.$$

If instead let $B$ be the point on the line $CA$ such that $A$ lies between $B$ and $C$, and that $$BA=\frac{\alpha\ CA}{2}$$ for some $\alpha>0$, then we can take $M$ to be the point between $A$ and $C$ such that $$AM=\frac{\alpha\ CA}{2(\alpha+1)}.$$  If $P$ and $Q$ are two distinct points on the circle with diameter $CA$ such that $P$, $M$, and $Q$ are collinear, then we have $$BP+BQ=(\alpha+1)\ PQ.$$
The proof is the same as the case $\alpha=1$.
