Question based of orthocenter distance from angular points

Question

In an acute angled triangle ABC,$\angle A=20^\circ $,let D,E,F be the feet of altitudes through A,B,C respectively and H is the orthocenter of $\bigtriangleup ABC $.Find $\frac{AH}{AD}+\frac{BH}{BE}+\frac{CH}{CF}$

Since $AH=2R cosA,AD=2R cos A+2R cos B cos C$

$\frac{AH}{AD}=\frac{2R cos A}{2R cos A+2R cosB cos C}=\frac{cos A}{cos A+cosB cos C}$

$BH=2R cosB,BE=2R cos B+2R cos A cos C$

$\frac{BH}{BE}=\frac{2R cos B}{2R cos B+2R cosA cos C}=\frac{cos B}{cos B+cosA cos C}$

$CH=2R cosC,CF=2R cos C+2R cos A cos B$

$\frac{CH}{CF}=\frac{2R cos C}{2R cos C+2R cosA cos B}=\frac{cos C}{cos C+cosA cos B}$

but since we have only A given,not B and C.How will we find these ratios?

Answer 1

The area $S$ of triangle $ABC$ is the sum of the areas of $ABH$, $BCH$ and $CAH$, so that: $$ S={1\over2}AB\cdot HF+{1\over2}BC\cdot HD+{1\over2}CA\cdot HE. $$

By dividing both sides by $S$ we get: $$ 1={AB\over2S} HF+{BC\over2S} HD+{CA\over2S} HE. $$ Observe now that $AB/2S=1/CF$, $BC/2S=1/AD$ and $CA/2S=1/BE$, so we may rewrite the above equality as: $$ 1={HF\over CF}+{HD\over AD} +{HE\over BE}. $$ Now plug in the obvious equalities $HF=CF-CH$, $HD=AD-AH$ and $HE=BE-BH$ to get:

$$ 1={CF-CH\over CF}+{AD-AH\over AD} +{BE-BH\over BE}, $$ that is: $$ 1=1-{CH\over CF}+1-{AH\over AD} +1-{BH\over BE} $$ and finally: $$ {CH\over CF}+{AH\over AD}+{BH\over BE}=2, $$ which is the sought-after result.

GOOD NEWS:

This result does not depend on the amplitude of the angles and holds for any point $H$ inside the triangle, provided $HD$, $HE$ and $HF$ are perpendicular to the sides of $ABC$. It is in fact a consequence of generalized Viviani's theorem.

BAD NEWS:

This result does not hold if $H$ is outside of the triangle (obtuse triangle). In that case however ${CH\over CF}+{AH\over AD}+{BH\over BE}$ does not have a fixed value, so the question cannot be answered.

Answer 2

Here's one solution using mass point geometry.

Assign mass points as following: $aA, bB, cC$ and let $(a+b+c)H$ be the center of the mass. Then from this it follows that points D, E, F have the following masses: $b+c, c+a, a+b$, respectively. Then from the mass points definition we have:

$$\frac{AH}{HD} = \frac{b+c}{a} \implies \frac{HD}{AH} + 1 = \frac{a}{b+c} + 1 \implies \frac{AD}{AH} = \frac{b+c+a}{b+c} \implies \frac{AH}{AD} = \frac{b+c}{a+b+c}$$

Simularly:

$$\frac{CH}{CF} = \frac{a+b}{a+b+c} \quad \text{and} \quad \frac{BH}{BE} = \frac{a+c}{a+b+c}$$

Adding them we get:

$$\frac{AH}{AD} + \frac{CH}{CF} + \frac{BH}{BE} = \frac{2(a+b+c)}{a+b+c} = 2$$

The fact that one of the angle is $20$ degrees is redundant.

If you're new to mass point geometry you can easily prove this by using Menelaus and Ceva Theorem. Actually the mass point geometry is an implicit and quicker way of using the Menalaus and Ceva Theorem, because if you want to use them explicitly, you need to find the right combination of lines and triangles.