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?

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.
