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If the equal sides AB and AC (each equal to 5 units) of a right-angled isosceles triangle ABC are produced to P and Q such that BP⋅CQ=AB², then the line PQ always passes through the fixed point (where A is the origin and AB, AC lie along the positive x and positive y - axis respectively)

User Ichiban
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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.

User Danexxtone
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