angle chasing, isosceles, <CBP=<BAP, P in median AM

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angle chasing, isosceles, <CBP=<BAP, P in median AM

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Source: All-Siberian Open School Olympiad 2018-19 9.4

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#1 h Jul 6, 2020, 6:53 AM • 1 Y

Y by Mango247

On the extension of the median $AM$ of an isosceles triangle $ABC$ with base $AC$ , take point $P$ such that $\angle CBP=\angle BAP$ . Find the angle $ACP$ .

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#2 h Jul 6, 2020, 2:08 PM

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My solution.

Extend line $BP$ and $AC$ , so that their intersection point is $E$ .
Since $ABC$ is an isosceles triangle, so $AB=BC$ .
Also, we see that $\triangle BPA \sim \triangle MBP$ , because $\angle CBP=\angle BAP$ and they have common angle $\angle BPM$ .
Thus, $\angle ABP=\angle BMP=\angle AMC$ and because $\angle BCA=\angle BAC$ , then $\triangle MCA \sim \triangle BAE$ and we denote that scale factor is $2$ ( $\frac{AB}{MC}=2$ ).
So, $AC=CE$ , $\angle PAE=\angle PEA$ and furthermore $AP=PE$ , hence $APE$ is an isosceles triangle and because $AC=CE$ we get that $PC$ is not only a median of $\triangle APE$ , it is the angle bisector of $\angle APE$ and it is also altitude, so $\angle ACP=90^\circ$ .

$[asy]size(14cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.1581139444696955, xmax = 8.841059277130235, ymin = -2.806932211481267, ymax = 9.416170967038065; /* image dimensions */ pen qqwuqq = rgb(0,0.39215686274509803,0); draw(arc((-2.04,0),0.5291386657367678,49.023352319033336,73.75236842721436)--(-2.04,0)--cycle, linewidth(2) + qqwuqq); draw(arc((0,7),0.5291386657367678,-74.05460409907715,-49.32460409907715)--(0,7)--cycle, linewidth(2) + qqwuqq); /* draw figures */ draw((-2.04,0)--(0,7), linewidth(2)); draw((0,7)--(2,0), linewidth(2)); draw((2,0)--(-2.04,0), linewidth(2)); draw((xmin, 1.1513157894736843*xmin + 2.348684210526316)--(xmax, 1.1513157894736843*xmax + 2.348684210526316), linewidth(2)); /* line */ draw((2,0)--(2.0092654850473983,4.661983288705887), linewidth(2)); draw((0,7)--(0,0), linewidth(2)); draw((xmin, -1.1636176148414552*xmin + 7)--(xmax, -1.1636176148414552*xmax + 7), linewidth(2)); /* line */ draw((2,0)--(6.015721926789293,0), linewidth(2)); /* dots and labels */ dot((-2.04,0),dotstyle); label("$A$", (-1.77698994865424,-0.33761843804301805), NE * labelscalefactor); dot((0,7),dotstyle); label("$B$", (0.07499538142444735,7.176150615419083), NE * labelscalefactor); dot((2,0),dotstyle); label("$C$", (2.068084355699606,0.17388227216919067), NE * labelscalefactor); dot((1,3.5),linewidth(4pt) + dotstyle); label("$M$", (1.0627208907997474,3.648559510507298), NE * labelscalefactor); dot((2.0092654850473983,4.661983288705887),linewidth(4pt) + dotstyle); label("$P$", (2.085722311224165,4.795026619603628), NE * labelscalefactor); dot((0,0),dotstyle); label("$D$", (0.07499538142444735,0.17388227216919067), NE * labelscalefactor); dot((6.015721926789293,0),linewidth(4pt) + dotstyle); label("$E$", (6.089538215299041,0.13860636112007282), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy]$

This post has been edited 2 times. Last edited by rafaello, Jul 6, 2020, 6:56 PM

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#3 h Jul 6, 2020, 2:50 PM

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$[asy] import geometry; pair A = dir(230), B = dir(90), C = dir(310), M = (B+C)/2, D = B+C-A; pair P = intersectionpoint(line(A, D), bisector(B, D)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(-10)); dot("$P$", P, dir(80)); draw(A--B--C--cycle); draw(B--D--C, dashed); draw(circumcircle(A,B,M), dotted); draw(A--D^^B--P); draw(anglemark(P,A,B)^^anglemark(M,B,P)^^anglemark(A,D,C), blue); [/asy]$
Let $D$ be the reflection of $A$ in $M$ so that $ABDC$ is a parallelogram. Since $\angle MAB = \angle MBP = \angle ADC$ and $\triangle BDC$ is isosceles, it follows that $CP$ is perpendicular to $BD$ and hence is also perpendicular to $AC$ .

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