Problem 3

Let

D

be an interior point of the acute triangle

ABC

with

AB>AC

so that

\angle DAB=\angle CAD

. The point

E

on the segment

AC

satisfies

\angle ADE=\angle BCD

, the point

F

on the segment

AB

satisfies

\angle FDA=\angle DBC

, and the point

X

on the line

AC

satisfies

CX=BX

. Let

O_1

and

O_2

be the circumcenters of the triangles

ADC

and

EXD

, respectively. Prove that the lines

BC

EF

, and

O_1O_2

are concurrent.

Step 5 of 6: Show the three centers are collinear

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In plain words

The common radical-center power forces the two original centers onto the perpendicular bisector through $P$ .

P,O_1,O_2\text{ are collinear}

Detailed analysis

Let $T$ be the second intersection of $(ACD)$ and $(EXD)$ . The inverse-circle construction gives $T$ on the circle centered at $P$ with radius $PD$ , so $PT=PD$ . Since $O_1$ and $O_2$ are the centers of circles through $D,T$ , both lie on the perpendicular bisector of $DT$ ; the same is true of $P$ . Hence $P,O_1,O_2$ are collinear.