Problem 2

Let

ABC

be a right triangle with

\angle B=90^\circ

. Point

D

lies on the line

CB

such that

B

is between

D

and

C

. Let

E

be the midpoint of

AD

and let

F

be the second intersection point of the circumcircle of

\triangle ACD

and the circumcircle of

\triangle BDE

. Prove that as

D

varies, the line

EF

passes through a fixed point.

Step 4 of 5: Solve for the second intersection F

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F=\left(\frac{c\left(d^2-a^2-2cd\right)}{a^2+d^2+4c^2-4cd},\ \frac{2ac(c-d)}{a^2+d^2+4c^2-4cd}\right)

Detailed analysis

Solving the two circle equations simultaneously gives exactly two solutions: $D=(-d,0)$ , and the point $F$ shown above (obtained by eliminating $x^2+y^2$ between the two equations and substituting back).