Problem
GEO-B1-M05-P022 Angle Between Diagonals
#22
★★★★☆ Level 4 of 5
In cyclic quadrilateral \(ABCD\), diagonals \(AC\) and \(BD\) meet at point \(P\). Prove that \(\angle APD=\angle BAC+\angle ABD\).
Consider triangle \(ABP\) and the exterior angle at point \(P\).
Since \(P\) lies on \(AC\), angle \(\angle BAP=\angle BAC\). Since \(P\) lies on \(BD\), angle \(\angle ABP=\angle ABD\). Angle \(\angle APD\) is an exterior angle of triangle \(ABP\) at vertex \(P\). Therefore it equals the sum of the two remote interior angles: \(\angle APD=\angle BAP+\angle ABP=\angle BAC+\angle ABD\).
The circle is not used directly in the equality, but the configuration is typical for later circular angle chasing problems.