In rectangle \(ABCD\) below, we have \(DP = PC\) and \(\angle BAC = 28^\circ\) . If \(Q\) is the point on \(\overline{AC}\) such that \(\overline{DQ}\perp \overline{AC}\), what is \(\angle QDP\) in degrees?
+128475 BAC = 28° = DCP [ a consequence of parallels AB and CD cut by transversal AC ]
And since DP = PC then angle DCP = angle PDC
Therefore angle DPC = 180 - 2(28) = 180 - 56 = 124
And by the exterior angle theorem
Angle QPD + Angle QDP = Angle DPC
90 + Angle QDP = 124
Angle QDP = 124 - 90 = 34°
