Problem: The sides of a triangle are of lengths 13,14, and 15. The altitudes of the triangle meet at point H. If AD is the altitude to side of length 14, the ratio HD:HA is:
Answer Choices:
A. 3:11
B. 5:11
C. 1:2
D. 2:3
E. 25:33
Solution:
142−q2=132−p2,27=q2−p2,15=q+p∴q=542,p=533
132−r2=152−s2,56=s2−r2,14=s+r∴s=9,r=5
AD2=132−52,AD=12,BE2=132−(533)2,BE=556
△HDB∼△HEA∴ut=pr=12−t556−u
Since r=5,p=533,u=2533t,12−t=12533⋅156−25⋅2533⋅33⋅t
∴t=116435,12−t=116957∴HD:HA=435:957=5:11
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