Loading DocumentIn a triangle ABC, point H is an orthocenter where the foot of the altitude B and C are D and E respectively. Suppose there is a circle with DE as diameter intersects triangle ABC on F and G, find the value of AK.
Solution :
We have BC = 25, BE = 7, BD = 20
⋅Angle CEB = Angle KFA ( 90°)
⋅Angle FAK = Angle ECB = 90° - <B
Therefore ∆AFK ~ ∆CEB
CBAK=CEAF ⇔AK=2425⋅AF
CD=√252−202=15
∆ADB ~ ∆AEC (AA) ⇔ AEAD=65=AD+15AE+7
{ 5AE=6AD6AE+42=5AD+75
⇒AE=11198=18⇒AD=65(18)=15
∆AFD ~ ∆AEC (AA)
AEAF=ACAD⇔AF=15+1515⋅18=9
Hence AK=2425⋅9=9.325