In triangle ABC, angle B = 90 degrees. Point X is on line AC such that angle BXA = 90 degrees, AX = 15, and CX = 5. What is BX?
A
15
X
5
B C
ABC ≈ BXC
ABC ≈ AXB
So
BXC ≈ AXB
BX / CX = AX / BX
BX^2 = AX * CX
BX^2 = 15 * 5
BX^2 = 75
BX = sqrt (75) = 5sqrt (3)
+45 
