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 1 of 4: Choose a parallel chord
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In plain words

Turn the angle condition into a central angle.

AD∥BC,∠ABD≥30∘,∠AOD≥60∘AD\parallel BC,\quad\angle ABD\ge30^\circ,\quad\angle AOD\ge60^\circ
Detailed analysis

Choose D D on the circumcircle with AD∥BC AD\parallel BC . Then ∠CBD=∠BCA\angle CBD=\angle BCA , so ∠ABD≥30∘\angle ABD\ge30^\circ and ∠AOD≥60∘\angle AOD\ge60^\circ .