Question

parallelogram abcd is divided into 4 triangles. △aed and △bcf are congruent isosceles triangles. △abf and △ecd are congruent right triangles. if angle ∠edc measures 50°, what is the measure of ∠fbc? a 30° b 45°

Explanation:

Step1: Recall properties of parallelogram

In parallelogram \(ABCD\), \(\angle ADC+\angle BCD = 180^{\circ}\) (adjacent - angles of a parallelogram are supplementary).

Step2: Analyze right - triangle \(\triangle EDC\)

In right - triangle \(\triangle EDC\), \(\angle DEC = 90^{\circ}\), and \(\angle EDC=50^{\circ}\). Then \(\angle ECD=180^{\circ}-\angle DEC - \angle EDC=180^{\circ}-90^{\circ}-50^{\circ}=40^{\circ}\).

Step3: Use congruence of triangles

Since \(\triangle ABF\cong\triangle ECD\), \(\angle ABF=\angle ECD = 40^{\circ}\). Also, since \(\triangle AED\cong\triangle BCF\) are isosceles triangles. In right - triangle \(\triangle BCF\), \(\angle BFC = 90^{\circ}\).
In right - triangle \(\triangle BCF\), we know that \(\angle FBC+\angle BCF = 90^{\circ}\). And \(\angle BCF=\angle ECD = 40^{\circ}\) (from congruence of \(\triangle ABF\) and \(\triangle ECD\)). So \(\angle FBC=40^{\circ}\).

Answer:

\(40^{\circ}\)