Problem Statement

Let ABC be a triangle with AB < AC < BC. Let the incentre and incircle of triangle ABC be I and ω, respectively. Let X be the point on line BC different from C such that the line through X parallel to AC is tangent to ω. Similarly, let Y be the point on line BC different from B such that the line through Y parallel to AB is tangent to ω. Let AI intersect the circumcircle of triangle ABC again at P ≠ A. Let K and L be the midpoints of AC and AB, respectively. Prove that ∠KIL + ∠YPX = 180°.

KolegaAI Solution

Proof:

Let's approach this proof using a combination of triangle geometry, circle geometry, and some of the concepts we've explored.

Lemma 1: In any triangle ABC with incentre I,

$$ ∠BIC = 90° + A/2 $$

where A is the measure of angle BAC.

Proof of Lemma 1:

  1. The incentre I is the point where the angle bisectors of the triangle meet.

  2. Therefore

    $$ ∠BAI = ∠CAI = A/2 $$

  3. In triangle BIC:

    $$ ∠BIC + ∠CBI + ∠BCI = 180° $$

    (angle sum in a triangle)

  4. But

    $$ ∠CBI = 90° + B/2 $$

    and

    $$ ∠BCI = 90° + C/2 $$

    (properties of the incentre)

  5. Therefore,

    $$ ∠BIC + (90° + B/2) + (90° + C/2) = 180° $$

  6. Simplifying:

    $$ ∠BIC + 180° + (B+C)/2 = 180° $$

$$
∠BIC = (B+C)/2
$$
  1. But

    $$ A + B + C = 180° $$

    (angle sum in a triangle), so

    $$ (B+C)/2 = 90° + A/2 $$

  2. Therefore,

    $$ ∠BIC = 90° + A/2 $$