The line PQ always passes through a fixed point.
Let A be the origin, B lie on the positive x-axis, and C lie on the positive y-axis such that AB = AC = 5. Let P and Q be points on the extended lines BA and CA, respectively, such that BP * CQ = AB^2. We need to prove that the line PQ always passes through a fixed point.
Consider the triangle BPC, where PC = AB = 5. Since BP * CQ = AB^2, we can write CQ = AB^2 / BP.
Using the Pythagorean theorem in triangle BPC, we get:
BC^2 = PC^2 + BP^2
Substituting PC = 5 and CQ = AB^2 / BP, we get:
BC^2 = 25 + (AB^2)^2 / (BP^2)
Moving the constants to one side and rearranging terms, we get:
(BP^2) * BC^2 - (AB^2)^2 = 25
Let F be the foot of the perpendicular drawn from B to AC, and let D be the midpoint of AC. Since triangle ABC is right-angled and isosceles, D is the center of the circle passing through points A, B, and C.
Draw lines DF and EF, forming right triangles BDF and BEF. Since D is the center of the circle, DF = DE = DC = 5/2.
In triangle BEF, using the Pythagorean theorem, we get:
BF^2 = BE^2 + EF^2
Substituting BE = AB = 5 and EF = (BC^2 - 5^2)^(1/2), we get:
BF^2 = 25 + (BC^2 - 25)^2
In triangle BDF, using the Pythagorean theorem, we get:
BD^2 = BF^2 + FD^2
Substituting BD = 5/2 and FD = 5/2, we get:
(5/2)^2 = BF^2 + (5/2)^2
Solving for BF^2, we get:
BF^2 = 12.5
Substituting BF^2 = 12.5 in the equation for BF^2 in triangle BEF, we get:
12.5 = 25 + (BC^2 - 25)^2
Solving for BC^2, we get:
BC^2 = 12.5
Substituting BC^2 = 12.5 in the equation for CQ, we get:
CQ = AB^2 / BP = 25 / BP
Now consider the equation (BP^2) * BC^2 - (AB^2)^2 = 25. Substituting BC^2 = 12.5 and AB^2 = 25, we get:
(BP^2) * 12.5 - (25)^2 = 25
Solving for BP^2, we get:
BP^2 = 7.5
Therefore, BP = √7.5.
This shows that for any given value of BP, the corresponding value of CQ is uniquely determined. This means that the line PQ always passes through a fixed point, namely the point where BP = √7.5.