Let $ABCD$ be a square. Let $M$ and $K$ be points on segments $BC$ and $CD$ respectively, such that $MC = KD$. Let $ P$ be the intersection of the segments $MD$ and $BK$. Prove that $AP$ is perpendicular to $MK$.
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Tags: perpendicular, square, geometry
Desimathematics
16.09.2021 18:05
It suffices to Show that $P$ is the orthocenter of triangle $AMK$ Triangle $ADK$ and $DMC$ are similar , this gives $MD \perp AK$ triangle $ABM$ and $BKC$ are similar this gives $BK \perp AM$ Done
e61442289
16.09.2021 18:33
parmenides51 wrote: Let $ABCD$ be a square. Let $M$ and $K$ be points on segments $BC$ and $CD$ respectively, such that $MC = KD$. Let $ P$ be the intersection of the segments $MD$ and $BK$. Prove that $AP$ is perpendicular to $MK$. Note that the triangle ABM is congruent to the triangle BKC by the LLL criterion (a - a + b - √a² + (a + b) ²), then <BAM = <KBC, then BK is perpendicular to AM, analogously we have that the triangle MDC is congruent to the triangle KAD by the criterion LLL (b - a + b - √b² + (a + b) ²), then <KAD = <MDC, then AK is perpendicular to MD, then P is orthocenter of AMK triangle, finally we have that AP is perpendicular to MK
OlympusHero
16.09.2021 20:03
Thought process: 1. draws diagram 2. darn looks pretty hard to do synthetically 3. oh looks pretty easy with coordinates 4. coordbash
Denote $A=(0,0), D=(a,0), C=(a,a), B=(0,a)$. Take $K=(a,b)$ and $M=(a-b,a)$. The equation of line $MD$ is $y=-\frac{a}{b}x+\frac{a^2}{b}$, and the equation of line $BK$ is $y=\frac{b-a}{a}x+a$, so the intersection is at $P=\left(\frac{a^3-a^2b}{a^2-ab+b^2}, \frac{a^2b}{a^2-ab+b^2}\right)$. The slope of the line $AP$ is hence $\frac{b}{a-b}$, and the slope of the line $MK$ is $\frac{b-a}{b}$. This means the two lines are perpendicular, as desired.