Problem
GEO-B1-M02-P021 The Diagonal of a Kite
In convex quadrilateral \(ABCD\), it is known that \(AB=AD\) and \(CB=CD\). Diagonals \(AC\) and \(BD\) meet at \(O\). Prove that \(BO=DO\) and \(AC\perp BD\).
First prove that \(AC\) bisects angle \(A\), then compare triangles \(ABO\) and \(ADO\).
Consider triangles \(ABC\) and \(ADC\). They have \(AB=AD\), \(CB=CD\), and common side \(AC\), so they are congruent by SSS. Therefore \(\angle BAC=\angle CAD\). Since \(O\) lies on \(AC\), we have \(\angle BAO=\angle OAD\). In triangles \(ABO\) and \(ADO\), \(AB=AD\), \(AO\) is common, and \(\angle BAO=\angle OAD\). By SAS, these triangles are congruent. Hence \(BO=DO\) and \(\angle AOB=\angle AOD\). Angles \(AOB\) and \(AOD\) are adjacent, so each is \(90^\circ\). Therefore \(AC\perp BD\).
Level 4 because it requires two consecutive uses of congruence and a perpendicularity conclusion.