Question

challenge! in the diagram below, de runs parallel to bc. m∠c = 26°, m∠a = 82°, and df bisects ∠bde as shown... what is the measure of angle dfb? 36° 54° 72° 82°

Explanation:

Step1: Find angle B

In $\triangle ABC$, use the angle - sum property of a triangle ($\angle A+\angle B+\angle C = 180^{\circ}$).
$m\angle B=180^{\circ}-m\angle A - m\angle C=180^{\circ}-82^{\circ}-26^{\circ}=72^{\circ}$

Step2: Use the parallel - line property

Since $DE\parallel BC$, $\angle BDE+\angle B = 180^{\circ}$ (same - side interior angles). So $m\angle BDE = 180^{\circ}-72^{\circ}=108^{\circ}$

Step3: Find angle BDF

Since $DF$ bisects $\angle BDE$, $m\angle BDF=\frac{1}{2}m\angle BDE=\frac{1}{2}\times108^{\circ}=54^{\circ}$

Step4: Find angle DFB

In $\triangle BDF$, use the angle - sum property of a triangle. Let $\angle DFB = x$. Then $x + m\angle BDF+m\angle B=180^{\circ}$. Substitute the known values: $x + 54^{\circ}+72^{\circ}=180^{\circ}$. Solving for $x$, we get $x = 54^{\circ}$

Answer:

$54^{\circ}$