AF _|_ BE wanted, in square ABCD (2020 Moldova NMO 7.7)

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AF _|_ BE wanted, in square ABCD (2020 Moldova NMO 7.7)

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#1 h Mar 17, 2021, 8:39 PM

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Let $ABCD$ be a square, and the point $E$ is the midpoint of the side $AD$ . If $BD \cap CE =\{F\}$ , prove that $AF \perp BE$

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#2 h Mar 17, 2021, 8:57 PM

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synthetic solution

This post has been edited 1 time. Last edited by natmath, Mar 17, 2021, 8:57 PM

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#3 h Mar 17, 2021, 8:58 PM

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@#2

Forgot a dollar sign there

Anyway I have better things to do than correcting latex on the forums so bye

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#4 h Mar 18, 2021, 12:41 AM • 1 Y

Y by ostriches88

Obviously we coord bash.

Set $A=(0,0),B=(0,6),C=(6,6),D=(6,0),E=(3,0)$ . Then with equations of lines, $F=(4,2)$ . Line $BE$ has slope $-2$ and line $AF$ has slope $\frac12$ , implying the result.

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#5 h Mar 18, 2021, 5:18 PM • 1 Y

Y by OlympusHero

Let $A=(0,0)$ , $b=(0,a)$ , $c=(a,a)$ , $d=(a,0)$ . We have $E=(\frac{a}{2},0)$ so the equation of line $BE$ is $y=-2x+a$ . Also the equation of line $CE$ is $y=2x-a$ . Equation of $BD$ is $y=-x+a$ so intersection with $CE$ is $-x+a=2x-a$ , meaning that $2a=3x$ or $x=\frac{2}{3}a$ and $y=\frac{1}{3}a$ . Therefore $F=(\frac{2}{3}a, \frac{1}{3}a)$ . Equation of $AF$ is $y=\frac{1}{2}x$ . Equation of $BE$ had slope $-2$ which is negative reciprocal of $AF$ . Therefore $\boxed{AF \perp BE}$ as desired. $\blacksquare$

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