Cute Since $Q$ lies on segment $\overline{MB}$, we have $CQ\geq QB$ (project onto $CB$), whence $PQ\geq QB$. Then \[ AP+PM=MQ+QB\leq MQ+PQ=2MQ+PM, \]so $AP\leq 2MQ$.

$\color{blue} \boxed{\textbf{SOLUTION 1}}$ Let, $\angle BAC=\beta, \angle ACP=\alpha \implies \angle CPQ=\angle PCQ=\alpha+\beta$ We have, $AM=BM=CM \implies \angle ACM=\angle MAC=\beta$ $\implies \angle PCM=\beta - \alpha \implies \angle MCQ=2\alpha$ And $\angle CQP=180°-2(\alpha+\beta)$ Using $\color{red} \textbf{Sine Law}$ on $\triangle APC,$ $\frac {AP}{sin\alpha}=\frac{AC}{sin(\alpha+\beta)}=\frac{ABcos\beta}{sin(\alpha+\beta)}...(1)$ Using $\color{red} \textbf{Sine Law}$ on $\triangle APC,$ $\frac {MQ}{sin2\alpha}=\frac{CM}{sin2(\alpha+\beta)}=\frac{AB}{2sin2(\alpha+\beta)}$ $\implies$ $\frac {MQ}{2sin\alpha cos\alpha}=\frac{AB}{2sin2(\alpha+\beta)}=\frac{AB}{4sin(\alpha+\beta) cos(\alpha+\beta)}...(2)$ $\frac{(2)}{(1)} \implies AP=2MQ [\frac{cos\beta cos(\alpha+\beta)}{cos\alpha}]$ $\textbf{Note :}$ $\alpha,\beta < 90°$ Now, $\frac{cos\beta cos(\alpha+\beta)}{cos\alpha}=cos^{2}\beta - tan\alpha sin\beta=1-(sin^{2} \beta +tan\alpha sin\beta) \leq 1$ Which gives $AP=2MQ [\frac{cos\beta cos(\alpha+\beta)}{cos\alpha}] \leq 2MQ \blacksquare$

$\color{blue} \boxed{\textbf{SOLUTION 2}}$ Let, $\angle BAC=\beta, \angle ACP=\alpha \implies \angle CPQ=\angle PCQ=\alpha+\beta$ We have, $AM=BM=CM \implies \angle ACM=\angle MAC=\beta$ $\implies \angle PCM=\beta - \alpha \implies \angle MCQ=2\alpha$ And $\angle CQP=180°-2(\alpha+\beta)$ Now, $\angle BCQ=90°-\beta-2\alpha \leq 90°-\beta=\angle CBQ \implies QB \leq CQ \implies QB \leq PQ$ So, $AP=AM-PM=BM-PM=MQ+QB-PM \leq MQ+PQ-PM=MQ+PM+MQ-PM=2MQ \blacksquare$

kfas wrote: Point $M$ is the midpoint of the hypotenuse $AB$ of a right angled triangle $ABC$. Points $P$ and $Q$ lie on segments $AM$ and $MB$ respectively and $PQ=CQ$. Prove that $AP\leq 2\cdot MQ$. Simply, we request $AQ-CQ\leq 2\left(AQ-\frac{AB}{2}\right)$; we write $AQ+CQ\geq AB$ equivalent. But surely, this is equivalent to $CQ\geq BQ$, for which is obvious.