Question

1. point p is on side \\(\overline{bc}\\) of rectangle abcd. if \\(ab = bp\\), find \\(m\angle apc\\).

Explanation:

Step1: Define rectangle properties

In rectangle \(ABCD\), \(AB = CD\), \(AD = BC\), all angles are \(90^\circ\), so \(\angle ABC = 90^\circ\). Let \(AB = BP = x\).

Step2: Find \(\angle BAP\)

In \(\triangle ABP\), \(AB = BP\) and \(\angle ABC = 90^\circ\), so it is an isosceles right triangle. Thus, \(\angle BAP = \angle BPA = \frac{180^\circ - 90^\circ}{2} = 45^\circ\).

Step3: Calculate \(\angle APC\)

\(\angle BPA\) and \(\angle APC\) are supplementary (they form a straight line on \(BC\)).
\(\angle APC = 180^\circ - \angle BPA = 180^\circ - 45^\circ\)

Answer:

\(135^\circ\)