Problem 4

Let triangle

ABC

with incenter

I

satisfy

AB<AC<BC

. Let

X

be a point on line

BC

, different from

C

, such that the line through

X

parallel to

AC

is tangent to the incircle. Similarly, let

Y

be a point on line

BC

, different from

B

, such that the line through

Y

parallel to

AB

is tangent to the incircle. Line

AI

intersects the circumcircle of triangle

ABC

again at

P

. Let

K

and

L

be the midpoints of

AC

and

AB

, respectively. Prove that

\angle KIL+\angle YPX=180^\circ

Step 3 of 4: T lies on line AP, giving two cyclic quadrilaterals

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In plain words

Since T is the reflection of A over I, it automatically lies on the same line AI that produces P on the circumcircle, so any angle at P towards A is secretly also an angle at P towards T; combined with the parallel line TY, this chases into a clean cyclic quadrilateral.

A,I,T,P\ \text{collinear};\qquad \angle TYC=\angle ABC=\angle APC=\angle TPC

Detailed analysis

Since $T=2I-A$ lies on line $AI$ , and $P$ is by definition the second intersection of line $AI$ with the circumcircle, the four points $A,I,T,P$ are collinear, so line $PT$ is the same as line $PA$ . Since $B,Y,C$ are collinear and $TY\parallel AB$ (Step 2), $\angle TYC=\angle TYB=\angle ABC$ (directed angles, using that line $BC$ is a single line through $Y$ ). Since $ABPC$ is cyclic and $B,P$ lie on the same arc relative to chord $AC$ , $\angle ABC=\angle APC$ ; and because $T$ lies on line $PA$ , $\angle APC=\angle TPC$ . Hence $\angle TYC=\angle TPC$ , so $T,Y,P,C$ are concyclic. The symmetric argument using $TX\parallel AC$ gives $\angle TXB=\angle ACB=\angle APB=\angle TPB$ , so $T,X,P,B$ are concyclic as well.