triangle Q?

Question

In triangle ABC,altitude from A to BC meets BC at D,and the altitude from B TO CA meets AD at H. AD=4,BD=3 CD=2,find HA

Answer 1

ACD is a right triangle, CD = 2, AD = 4, so AC = 2 sqrt [5]

ABD is a right triangle, BD = 3, AD = 4, so AB = 5.

BC = BD + CD = 5

ABC is an isosceles triangle, so BH bisects AC.

AH = 1/2 AC = sqrt[5]

Answer 2

BH meets AC at E -> BE is perpendicular to AC at E
=> angel EBC = angel CAD ( BE is perpendicular to AC , BC is perpendicular to AD )
=> tri. ADC and tri. BDH is similar
=> BD / AD = DH / DC => DH = BD * DC / AD = 1.5
AH = AD - DH = 2.5