Let $ABCD$ be a square. On the sides $BC$ and $CD$ the points $M$ and $K$ respectively, so that $MC = KD$. Let $P$ the intersection point of of segments $MD$ and $BK$. Prove that $AP \perp MK$.
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Tags: geometry, square, perpendicular
vanstraelen
17.09.2024 22:34
Given the square $ABCD\ :\ A(0,0),B(1,0),C(1,1),D(0,1)$. Choose $M(1,\lambda)$, then $K(1-\lambda,1)$. The line $DM\ :\ y=1+(\lambda-1)x$ intersects the line $BK\ :\ y=-\frac{1}{\lambda}(x-1)$ in the point $P(\frac{1-\lambda}{\lambda^{2}-\lambda+1},\frac{\lambda}{\lambda^{2}-\lambda+1})$. Slope of the line $AP\ :\ m_{AP}=\frac{\lambda}{\lambda-1}$, slope of the line $MK\ :\ m_{MK}=\frac{\lambda-1}{\lambda}$, so $AP \bot MK$.
teomihai
18.09.2024 18:33
Tsikaloudakis wrote: see the figure: wow solution!