If $D$, $E$, $F$ are feet of perpendiculars from a point to the sides of $\triangle ABC$, then $BD^2-DC^2+CE^2-EA^2+AF^2-FB^2=0$

Question

The actual question is

If from a point $O$, segments $OD$, $OE$, $OF$ are drawn perpendicular to the sides $BC$, $CA$, $AB$, respectively, of $\triangle ABC$, then prove that: $$BD^2-DC^2+CE^2-EA^2+AF^2-FB^2=0$$

I know the question can be easily solved by applying Ceva's theorem.

I have drawn the $AD$, $CF$, $BE$ to sides $BC$, $BA$, $AC$ respectively, but assuming them to be concurrent . Then we can easily derive $$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$$ further squaring both side will give us the required result

My doubt is how to prove $AD,CF,BE$ to be concurrent .

I prove the result as part of this answer. I coined the term "orthians" for the perpendiculars at $D$, $E$, $F$, and back-constructed the name "Ortha's Theorem" for the condition of their concurrency. The official name, however, is Carnot's Theorem. — Blue, Jun 12 '19 at 07:19

score 2 · Accepted Answer · answered Jun 12 '19 at 07:08

2

It has nothing to do with the Ceva theorem.

Hint:

It is well know that $MN\bot XY$ iff $$MX^2-MY^2 = NX^2-NY^2$$

You can use that in order to finsih your problem. Say $$OB^2-OC^2 = DB^2-DC^2$$ $$...$$

answered Jun 12 '19 at 07:08

nonuser

90,026

If $D$, $E$, $F$ are feet of perpendiculars from a point to the sides of $\triangle ABC$, then $BD^2-DC^2+CE^2-EA^2+AF^2-FB^2=0$

1 Answers1