In a right triangle $ABC$ where the angle $ABC = 90^{\circ}$, on the side $BC$ is placed a point $P$ and on $AC$ a point $M$ so that $BM$ and $AC$ are perpendicular and $AP=PC=MC=1$. What would be the length of $BC$?
No plot is given so I plotted it in GeoGebra.
I drew the segment $MP$ to create the isosceles triangle $MPC$ but I did not get any result. I also tried to create the isosceles triangles $MPC$, $APC$, and the right triangles $BMC$, $BMA$ and $ABP$. I also placed a point at the intersection of $BM$ and $AP$ called $D$, with this point I created another right triangle $ADM$.
I really appreciate any help you can provide.
Best Answer
As you did, label the intersection point of $BM$ and $AP$ as $D$. Since $\lvert AP\rvert = \lvert PC\rvert = 1$, then $\triangle APC$ is isosceles and, thus,
$$\measuredangle PAC = \measuredangle PCA = \gamma \;\;\to\;\; \measuredangle APC = 180^{\circ} - 2\gamma, \; \measuredangle APB = 2\gamma \tag{1}\label{eq1A}$$
Also, let
$$\lvert AM\rvert = x, \;\; \lvert AD\rvert = y \;\;\to\;\; \lvert DP\rvert = 1 - y \tag{2}\label{eq2A}$$
In $\triangle ADM$, since $\measuredangle AMD = 90^{\circ}$, then $\measuredangle ADM = 90^{\circ} - \gamma$. Also, by opposite angles, we have $\measuredangle BDP = 90^{\circ} - \gamma$. Thus, from \eqref{eq1A}, we get
$$\measuredangle DBP = 180^{\circ} - \measuredangle BDP - \measuredangle APB = 180^{\circ} - (90^{\circ} - \gamma) - 2\gamma = 90^{\circ} - \gamma \tag{3}\label{eq3A}$$
Since $\measuredangle BDP = \measuredangle DBP$, then $\triangle BDP$ is isosceles, with using \eqref{eq2A} that $\lvert BP\rvert = \lvert DP\rvert = 1 - y$. This then gives that
$$\lvert BC \rvert = \lvert BP\rvert + \lvert PC\rvert = (1 - y) + 1 = 2 - y \tag{4}\label{eq4A}$$
Next, from the first part of \eqref{eq1A}, we get that $\triangle DAM \sim \triangle BCM$ so, by ratios of similar sides, we have
$$\frac{1}{2 - y} = \frac{x}{y} \;\;\to\;\; x = \frac{y}{2 - y} \tag{5}\label{eq5A}$$
Thus, then also
$$\lvert AC\rvert = \lvert AM\rvert + \lvert MC\rvert = \frac{y}{2 - y} + 1 = \frac{y + (2 - y)}{2 - y} = \frac{2}{2 - y} \tag{6}\label{eq6A}$$
Since $\triangle ACB \sim \triangle BCM$ then, from \eqref{eq2A}, \eqref{eq4A} and \eqref{eq5A} we get
$$\frac{\frac{2}{2-y}}{2 - y} = \frac{2 - y}{1} \;\;\to\;\; 2 = (2 - y)^3 \;\;\to\;\; 2 - y = \sqrt[3]{2} \tag{7}\label{eq7A}$$
Finally, using \eqref{eq4A}, we have
$$\lvert BC\rvert = \sqrt[3]{2} \tag{8}\label{eq8A}$$