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Let $ABC$ be a triangle with $AB=AC$. If $D$ is the midpoint of $BC$, $E$ is the foot of the perpendicular drawn from $D$ to $AC$ and $F$ the mid-point of $DE$, prove that $AF$ is perpendicular to $BE$.
(JEE-1989)

Figure

I have solved it using coordinate geometry (or rather truthfully, seen the solution), but was wondering if it could be solved by geometry.
Thanks in advance.

MvG
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Himanshu Singh
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  • What is “JEE-1989”? Can you spell that out and/or add a link? I've found several copies of this question on the web, but non in a book matching these initials. When you say “by geometry”, does that include trigonometry? – MvG Oct 31 '14 at 08:05
  • JEE stands for Joint Engineering Examination and 1989 is the year of the paper. It is a rather difficult competitive examination in which this question appeared and has no real relevance to the question. And i have no objections to you using trignometry. I just wanted to know if it was possible to solve it without using coordinate geometry. – Himanshu Singh Oct 31 '14 at 12:03
  • I'm trying to create a site for "Indian Competitive Exams" and your question seems to be related to it, I think you would be interested to support it here and spread the word to your friends? – RE60K Apr 25 '15 at 07:44

1 Answers1

2

Let $M$ be the midpoint of $BD$. Drop a perpendicular from $D$ to $AM$, meeting it at $P$. Drop a perpendicular from $D$ to $AB$, meeting it at $H$.

diagram

Part 1: $BPE$ are collinear.

Since $$ \frac{AP}{PM} = \frac{AP}{PD}\cdot\frac{PD}{PM} = \left(\frac{AD}{DM}\right)^2 $$ and similarly $\frac{CE}{EA} = \left(\frac{CD}{AD}\right)^2$, we have $$ \frac{AP}{PM}\cdot\frac{MB}{BC}\cdot\frac{CE}{EA} = \left(\frac{AD}{DM}\right)^2 \left(-\frac14\right) \left(\frac{CD}{AD}\right)^2 = \left(\frac{CD}{DM}\right)^2 \left(-\frac14\right) = -1 $$ Thus $BPE$ are collinear by Menelaus' theorem.

Part 2: $\angle DPE = \angle BAD$.

Since $DP\perp PA$ and $DH\perp HA$, by Thales' theorem $DPHA$ are concyclic, as shown. Therefore $\angle PHD = \angle PAD$, since they stand on the same arc $PD$. Now, since $\angle PHB = \frac\pi2-\angle PHD$, and $\angle PMB = \pi-\angle PMD = \pi-(\frac\pi2-\angle PAD)$, it follows that $\angle PHB+\angle PMB = \pi$, and so $BMPH$ are concyclic, as shown. Next, since $BH\perp HD$ and $M$ is the midpoint of $BD$, we have $MH=MB$; therefore $\angle MBH$ and $\angle BPM$ stand on equal arcs, and so they are equal angles. Finally, $$ \angle DPE = \tfrac\pi2 - \angle BPM = \tfrac\pi2-\angle MBH = \angle BAD $$

Part 3: $AF\perp BE$.

Apply the similarity transformation taking $\triangle ABD$ to $\triangle ADE$, that is, rotating about $A$ by $\angle BAD$ and then scaling by $\frac{AD}{AB}$. This map sends $D$ to $E$ and sends the median $AM$ to the median $AF$. Since we rotate by $\angle BAD = \angle DPE$, it sends the line $DP$ to a line parallel to $EB$, so it must send the line $DP$ to the line $EB$. Since $DP\perp AM$ by construction, this shows that $EB\perp AF$.

Remark: If we consider $A$ and $B$ to be fixed and let $C$ vary on a circle, the locus of point $E$ is a cardioid. Maybe some short proof is possible using the properties of cardioid...