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 3 of 6: Obtain the inversion pairs

Previous step Next step

In plain words

Equal powers identify pairs exchanged by one inversion.

PD^2=PC\cdot PB=PE\cdot PF

Detailed analysis

The tangent-secant theorem at $P$ gives $PD^2=PC\cdot PB$ . Since $PD$ is tangent to $(DEF)$ , it also gives $PD^2=PE\cdot PF$ . Thus inversion in the circle centered at $P$ with radius $PD$ exchanges $B\leftrightarrow C$ and $E\leftrightarrow F$ .