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°.
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:
The incentre I is the point where the angle bisectors of the triangle meet.
Therefore
$$ ∠BAI = ∠CAI = A/2 $$
In triangle BIC:
$$ ∠BIC + ∠CBI + ∠BCI = 180° $$
(angle sum in a triangle)
But
$$ ∠CBI = 90° + B/2 $$
and
$$ ∠BCI = 90° + C/2 $$
(properties of the incentre)
Therefore,
$$ ∠BIC + (90° + B/2) + (90° + C/2) = 180° $$
Simplifying:
$$ ∠BIC + 180° + (B+C)/2 = 180° $$
$$
∠BIC = (B+C)/2
$$
But
$$ A + B + C = 180° $$
(angle sum in a triangle), so
$$ (B+C)/2 = 90° + A/2 $$
Therefore,
$$ ∠BIC = 90° + A/2 $$