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#1 h Mar 20, 2025, 11:14 AM

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Let $ABC$ be a triangle. Let $E,F$ be the feet of the altitudes from $B,C$ respectively. Let $P,Q$ be the projections of $B,C$ onto line $EF$ . Show that $PE=QF$ .

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#3 h Mar 20, 2025, 12:46 PM • 2 Y

Y by MS_asdfgzxcvb, Lil_flip38

good one. Of course we have to use DDIT. Let $H=BE\cap CF$ be the orthocenter, $\infty$ be the point at infinity along the direction perpendicular to $EF$ , and $M$ the midpoint of $EF$ . The projections from $A$ and $H$ to $EF$ are $A’$ and $H’$ , respectively. As $AH$ is a diameter of $(AEFH)$ , $M$ is also the midpoint of $A’H’$ . By DDIT from $\infty$ to $BECF$ and projecting onto $EF$ , we find an involution swapping $(P,Q),(E,F),(A’,H’)$ . So $M$ is the midpoint of $PQ$ as well - i.e.e $PE=QF$ .

This post has been edited 1 time. Last edited by keglesnit, Mar 21, 2025, 8:08 PM

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#4 h Mar 20, 2025, 2:00 PM • 1 Y

Y by MrdiuryPeter

Trigonometry bash

Let $H$ be the orthocenter of $\triangle ABC$
It is well known fact that $HC=2RcosC$
$\therefore \frac{EC}{sinA}=HC=2RcosC \implies EC=2RcosCsinA$
applying sine rule to $\triangle QEC \implies \frac{QE}{cosB}=EC \implies QE=2RcosBcosCsinA$
so as $QE$ is symmetric is $\angle B$ and $\angle C \implies QE=PF$

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#5 h Mar 20, 2025, 2:17 PM

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Let $N,M$ be the midpoint of $EF,BC$ respectively. As $BCEF$ is cyclic with center $M$ , $MN\perp EF$ . Then we have $BP\parallel MN\parallel PQ$ and $M$ being the midpoint of $BC$ so $N$ has to be the midpoint of $PQ$ so $PE=QF$

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#6 h Mar 20, 2025, 11:54 PM

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Easy one.
Define: $\angle PBF = \beta$ , $\angle ECQ = \alpha$
$\leftrightarrow PF = EQ$
$\leftrightarrow CE.Cos(\alpha) = BF.Cos(\beta)$
$\leftrightarrow \frac{CE}{BF} = \frac{Cos(\beta)}{Cos(\alpha)}$
Which is true using sine's law.

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#7 h Mar 21, 2025, 8:10 AM

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SEE THE FIGURE

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#8 h Apr 12, 2025, 10:44 PM

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Let $M$ the midpoint of $BC$ , from perpendicular trivially notice $M$ is center of $(BFEC)$ which means that if we let $N$ midpoint of $EF$ then $MN \perp EF$ so it is in fact perpendicular bisector, however from thales we must also have $PN=QN$ and this is enough to finish thus we are done :cool:

:cool:

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