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 1 of 4: A homothety turns K, I, L into C, T, B

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In plain words

The homothety centered at A with ratio 2 sends the midpoint of a segment to its far endpoint, so it sends L to B, K to C, and I (the midpoint of A and its reflection T) to T itself, turning angle KIL directly into angle BTC.

\angle KIL=\angle BTC,\qquad T=2I-A

Detailed analysis

Let $T$ be the reflection of $A$ over $I$ , so $T=2I-A$ . The homothety centered at $A$ with ratio $2$ sends $L$ (midpoint of $AB$ ) to $B$ , sends $K$ (midpoint of $AC$ ) to $C$ , and sends $I$ to $2I-A=T$ . Since a homothety preserves angles, $\angle KIL=\angle CTB=\angle BTC$ .