1. ## geometry

Question
In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

Solution
In the triangle ABC, the sides are:
1. AB,
2. BC=1.5* AB
3. AC=2* AB
and CH+AH = AC.
COS(angle A) = AH / AB
COS(angle C) = CH / BC = CH / (1.5* AB)

CH / AH = (1.5* AB)*COS(angle C) / (AB * COS(angle A) )
CH / AH = (1.5)*COS(angle C) / (2 - COS(angle C) )
Hence, CH / AH = 1.90909...= 21 / 11

Is this the correct way to do it?

2. Originally Posted by xwrathbringerx
Question
In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

Solution
In the triangle ABC, the sides are:
1. AB,
2. BC=1.5* AB
3. AC=2* AB
and CH+AH = AC.
COS(angle A) = AH / AB
COS(angle C) = CH / BC = CH / (1.5* AB)

CH / AH = (1.5* AB)*COS(angle C) / (AB * COS(angle A) )
CH / AH = (1.5)*COS(angle C) / (2 - COS(angle C) )
Hence, CH / AH = 1.90909...= 21 / 11

Is this the correct way to do it?
Maybe I am still rusty, but how did you get
[(1.5)cos(angle C)] / [2 -1.5cos(angle C)] = 1.90909...?

3. Can anyone help me solve this problem?!?

4. Originally Posted by xwrathbringerx
Can anyone help me solve this problem?!?
Question
In a triangle ABC, AC = 2AB and BC/BA = 3/2. H is the foot of the perpendicular from B to AC. Prove that CH/AH = 21/11.

Let AB = u
Then, AC = 2u
And, BC = (3/2)u = 1.5u

Let also AH = v
So, HC = 2u -v

In right triangle AHB, by Pythagorean theorem,
(BH)^2 = (AB)^2 -(AH)^2 -------(i)

In right triangle BHC,
(BH)^2 = (BC)^2 -(HC)^2 ------(ii)

(BH)^2 = (BH)^2, so,
(AB)^2 -(AH)^2 = (BC)^2 -(HC)^2
Substitutions,
u^2 -v^2 = (1.5u)^2 -(2u -v)^2
u^2 -v^2 = 2.25u^2 -4u^2 +4uv -v^2
The -v^2 cancels out,
u^2 -2.25u^2 +4u^2 = 4uv
2.75u^2 = 4uv
2.75u = 4v
v = (2.75 /4)u = (5.5 /8)u = (11 /16)u -------**

Now,
(CH) / (AH)
= (2u -v) / v
= [2u -(11/16)u] / [(11/16)u]
= [(32u -11u) /16] / [11u /16]
= [21u /16] *[16 /(11u)]
= 21/11 -------------------------proven.