Question

1. in the figure given, ad is perpendicular and equal to ab. if dc = 2ab, what is the degree measure of angle abc? a) 125 b) 135 c) 145 d) 155

Explanation:

Step1: Extend \(AB\) and \(DC\) to meet at point \(E\)

Since \(AD\perp AB\) and \(AD\perp DC\) (as \(AD\) is perpendicular to \(AB\) and \(AB\parallel DC\) in the trapezoid - like figure), \(\angle ADE = 90^{\circ}\), \(\angle DAE=90^{\circ}\). Also, \(AD = AB\) (given \(AD\) is perpendicular and equal to \(AB\)), and \(DC = 2AB\). Let \(AB=x\), then \(AD=x\) and \(DC = 2x\), so \(EC=DC - DE\) (where \(DE = AB\) because of the rectangle \(ADEB\) formed). In rectangle \(ADEB\), \(DE=AB\) and \(BE = AD\). So \(EC=x\) and \(BE=x\).

Step2: Analyze \(\triangle BEC\)

In \(\triangle BEC\), \(BE = EC=x\) and \(\angle BEC = 90^{\circ}\) (since \(\angle ADE=\angle DAE = 90^{\circ}\)). So \(\angle EBC=\angle ECB = 45^{\circ}\) (using the property that in a right - isosceles triangle \(\triangle BEC\) (\(a^{2}+a^{2}=c^{2}\), and \(\angle BEC = 90^{\circ}\), \(\tan\angle EBC=\frac{EC}{BE}=1\)), \(\angle EBC = 45^{\circ}\)).

Step3: Calculate \(\angle ABC\)

\(\angle ABC=180^{\circ}-\angle EBC\) (linear - pair of angles).

Answer:

\(135^{\circ}\), so the answer is B.