MathLabs

Problem 1

Let ABC ABC be an acute-angled triangle with circumcentre O O . Let P P on BC BC be the foot of the altitude from A A . Suppose that ∠BCA≥∠ABC+30∘\angle BCA\ge\angle ABC+30^\circ . Prove that ∠CAB+∠COP<90∘\angle CAB+\angle COP<90^\circ .
Step 4 of 4: Finish the angle comparison
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In plain words

Use a diameter for complementary angles.

∠COP<∠OCP=90∘−∠CAB\angle COP<\angle OCP=90^\circ-\angle CAB
Detailed analysis

If CE CE is a diameter, then ∠OCP=∠ECB=∠EAB\angle OCP=\angle ECB=\angle EAB and ∠EAB+∠CAB=90∘\angle EAB+\angle CAB=90^\circ . Therefore ∠CAB+∠COP<90∘\angle CAB+\angle COP<90^\circ .