In triangle \(ABC, BC = 4, AC = 3\sqrt2,\) and \(\angle C = 45^\circ.\) Altitudes \(AD, BE,\) and \(CF\) intersect at the orthocenter \(H.\) Find \(AH:HD\)
+128475 Let A =(0, 0) Let B =(sqrt 10 , 0)
Using the Law of Cosines we can find AB as
AB^2 = BC^2 + AC^2 - 2(BC * AC) cos (C)
AB^2 = 16 + 18 - 2(4 * 3sqrt(2)) * 1 /sqrt (2)
AB^2 = 34 - 24
AB^2 = 10
AB = sqrt (10)
Area of triangle =(1/2) (4)(3sqrt 2) (1/sqrt 2) = 6
Also area of triangle = (1/2) BC * altitude drawn to AB
Call the altitude drawn to AB = CF ......so.....
6 = (1/2) sqrt (10) CF
CF = 12/sqrt (10)
AD^2 = AC^2 - CF^2
AD^2 = 18 - 144/10
AD^2 = 36/10
AD = 6 / sqrt 10
Equation of CF : x = 6/sqrt 10
So C = ( 6/sqrt 10 , 12/sqrt 10)
Slope of AC = ( 12/ sqrt 10 ) / ( 6 /sqrt 10) = 12/6 = 2
Slope of line perpendicular ro AC = -1/2
Equation of perpendicular line to AC : y = (-1/2)(x - sqrt 10) = -x/2 + sqrt (10) / 2
This line intersects AF at : y = (-6/sqrt 10)/2 + sqrt (10) / 2 = sqrt (10) / 5
So
H = ( 6/sqrt 10 , sqrt (10) / 5 )
Distance from A to H =sqrt [ 36/10 + 10/25 ] = 2
Slope of line BC = (12/sqrt 10) / (6/sqrt 10 - sqrt 10) = -3
Equation of line through BC : y = -3( x - sqrt 10) = -3x + 3sqrt 10
Standard form of this line
3x + y - 3sqrt 10 = 0
Distance from A to D =
abs ( 3(0) + 0 -3sqrt 10) / sqrt( 3^2 + 1^2) = 3sqrt 10 / sqrt 10 = 3 = AD
So
HD = AD - AH = 3 - 2 = 1
So
AH : AD = 2 : 1
See the image here :
