# Question based of orthocenter distance from angular points

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?

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.

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.

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:

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

$$\frac{AH}{AD} + \frac{CH}{CF} + \frac{BH}{BE} = \frac{2(a+b+c)}{a+b+c} = 2$$
The fact that one of the angle is $20$ degrees is redundant.