Question

point d is the incenter of triangle bca. if m∠fhg = 61°, what is the measure of ∠fdg?

Explanation:

Step1: Recall the property of the in - center and the circle

Since \(D\) is the in - center of \(\triangle BCA\), \(DF\perp BC\), \(DH\perp BH\), \(DG\perp CG\) (tangent - radius property: a radius is perpendicular to the tangent at the point of tangency). So \(DF = DH=DG\) and the quadrilateral \(DFHG\) is cyclic (a quadrilateral with \(DF\perp BF\), \(DG\perp CG\) and \(D\) is the in - center).

Step2: Use the property of angles in a cyclic quadrilateral

The sum of opposite angles in a cyclic quadrilateral is \(180^{\circ}\). In cyclic quadrilateral \(DFHG\), we know that \(\angle FHG+\angle FDG = 180^{\circ}\)

Step3: Solve for \(\angle FDG\)

Given \(m\angle FHG = 61^{\circ}\), then \(m\angle FDG=180^{\circ}-m\angle FHG\)

Substitute \(m\angle FHG = 61^{\circ}\) into the formula: \(m\angle FDG = 180^{\circ}-61^{\circ}\)

\(m\angle FDG=119^{\circ}\)

Answer:

\(119^{\circ}\)