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 5 of 5: The line EF always meets BC at the same point

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\text{line } EF \cap \{y=0\} = (-c,0),\ \text{independent of } d

Detailed analysis

Writing the line through $E$ and $F$ and setting $y=0$ gives the $x$ -intercept $x=-c$ , which does not depend on $d$ at all. Hence, as $D$ (equivalently $d$ ) varies, the line $EF$ always passes through the fixed point $P=(-c,0)$ , i.e. the point on ray $CB$ beyond $B$ with $BP=BC$ .