Problem
GEO-B2-M01-P009 Angle Between Diagonals
#9
★★★☆☆ Level 3 of 5
In convex cyclic quadrilateral \(ABCD\), diagonals \(AC\) and \(BD\) intersect at \(P\). Prove that \(\angle APB=\angle ACB+\angle CBD\).
Look at triangle \(BCP\), then use the supplementary angle at \(P\).
In triangle \(BCP\), we have \(\angle PBC=\angle DBC=\angle CBD\) and \(\angle BCP=\angle BCA=\angle ACB\). Therefore \(\angle BPC=180^\circ-\angle CBD-\angle ACB\). Angles \(\angle APB\) and \(\angle BPC\) are supplementary, hence \(\angle APB=\angle ACB+\angle CBD\).
A good problem on carefully moving between ordinary and oriented angles.