We consider triangle $ABC$ with $\angle BAC = 90^{\circ}$ and $\angle ABC = 60^{\circ}.$ Let $ D \in (AC) , E \in (AB),$ such that $CD = 2 \cdot DA$ and $DE $ is bisector of $\angle ADB.$ Denote by $M$ the intersection of $CE$ and $BD$, and by $P$ the intersection of $DE$ and $AM$. a) Show that $AM \perp BD$. b) Show that $3 \cdot PB = 2 \cdot CM$.
Problem
Source: Romania National Olympiad 2023
Tags: geometry, Metric Relation, perpendicularity
lpieleanu
01.07.2023 21:07
Without loss of generality, let $$A=(0,0),$$$$B=(0,6),$$$$C=(6\sqrt{3},0).$$Then, we have $$D = (2\sqrt{3},0).$$Note that $$\frac{AB}{AD} = \frac{6}{2\sqrt{3}} = \sqrt{3}$$which implies that $ABD$ is a $30^\circ-60^\circ-90^\circ$ triangle with right angle at $A$ and $30^\circ$ angle at $B.$ Thus, by the Angle Bisector Theorem, we have $$\frac{AE}{EB} = \frac{AD}{BD} = \frac{1}{2},$$so $$E = (0,2).$$Then, the line going through $E$ and $C$ has slope $\tfrac{2 - 0}{0-6\sqrt{3}} = -\tfrac{1}{3\sqrt{3}} = -\tfrac{\sqrt{3}}{9}$ and $y$-intercept $(0,2),$ so its equation is $$y = -\frac{\sqrt{3}}{9}x +2.$$Similarly, we can derive that the equation of the line containing $B$ and $D$ is $$y=-\sqrt{3}x + 6.$$Therefore, the intersection of the lines $y = -\frac{\sqrt{3}}{9}x +2$ and $y=-\sqrt{3}x + 6$ is $$M = \left( \frac{3\sqrt{3}}{2}, \frac{3}{2} \right).$$Then, we can find that the slope of the line going through $A$ and $M$ is $\tfrac{\tfrac{3}{2}}{\tfrac{3\sqrt{3}}{2}} = \tfrac{1}{\sqrt{3}} = \tfrac{\sqrt{3}}{3},$ while the slope of the line going through $B$ and $D$ is $\tfrac{0-6}{2\sqrt{3}-0} = -\sqrt{3}.$ Since these slopes are negative reciprocals, we can conclude that $AM \perp BD.$ $\square$
Let us use the same diagram as for part $(a).$ Since the slope of the line going through $A$ and $M$ is $\frac{\sqrt{3}}{3}$ and this line goes through the origin, the equation of this line is $$y = \frac{\sqrt{3}}{3}x.$$Recall that $$D=(2\sqrt{3},0),$$$$E = (0,2),$$so the slope of the line going through $D$ and $E$ is $\tfrac{2-0}{0-2\sqrt{3}} = -\tfrac{\sqrt{3}}{3}.$ Its $y$-intercept is $(0,2)$ so it follows that the equation of this line is $$y = -\frac{\sqrt{3}}{3}x + 2.$$The intersection of this line and line $AM$ is $$P=(\sqrt{3}, 1).$$Now, we simply use the distance formula to compute $$PB = \sqrt{(0-\sqrt{3})^2 + (6-1)^2} = \sqrt{28} = 2\sqrt{7},$$and $$CM = \sqrt{\left( 6\sqrt{3} - \frac{3\sqrt{3}}{2} \right)^2 + \left( 0-\frac{3}{2} \right)^2} = \sqrt{\frac{243}{4} + \frac{9}{4}} = \sqrt{63} = 3\sqrt{7} = \frac{3}{2} \cdot PB,$$so we are done. $\square$