Question
Point D is the incenter of triangle BCA. If m∠FDG = 136°, what is the measure of ∠FHG?
Triangle BCA with inscribed circle D. Segments BF and BH, CF and CG, and AG and GH are tangent to circle D; segments FD, GD, FH, and GH are created from points F, G, D, and H.
68°
44°
136°
88°
Triangle BCA with inscribed circle D. Segments BF and BH, CF and CG, and AG and GH are tangent to circle D; segments FD, GD, FH, and GH are created from points F, G, D, and H.
68°
44°
136°
88°
Answers
GPT-4o mini
To find the measure of angle \( \angle FHG \), we can use the properties of tangents to a circle.
Since \( D \) is the incenter of triangle \( BCA \), and segments \( BF, BH, CF, CG, AG, GH \) are all tangent to the circle inscribed at \( D \), we know that the angles formed between a tangent and a line from the center of the circle are equal to the angles opposite the tangent line.
Given that \( \angle FDG = 136^\circ \), we can find \( \angle FHG \).
1. The angles \( \angle FDH \) and \( \angle GDH \) are both equal to \( 90^\circ \) because they are angles between the tangent line segments (\( FH \) and \( GH \)) and the radii \( DF \) and \( DG \) of the circle at points of tangency to lines \( FH \) and \( GH \).
2. Since the angles at \( D \) are supplementary:
\[
\angle FDG + \angle FHG = 180^\circ
\]
3. Therefore, substituting the value we have:
\[
136^\circ + \angle FHG = 180^\circ
\]
4. We can subtract \( 136^\circ \) from \( 180^\circ \):
\[
\angle FHG = 180^\circ - 136^\circ
\]
\[
\angle FHG = 44^\circ
\]
Thus, the measure of \( \angle FHG \) is \( \boxed{44^\circ} \).
Since \( D \) is the incenter of triangle \( BCA \), and segments \( BF, BH, CF, CG, AG, GH \) are all tangent to the circle inscribed at \( D \), we know that the angles formed between a tangent and a line from the center of the circle are equal to the angles opposite the tangent line.
Given that \( \angle FDG = 136^\circ \), we can find \( \angle FHG \).
1. The angles \( \angle FDH \) and \( \angle GDH \) are both equal to \( 90^\circ \) because they are angles between the tangent line segments (\( FH \) and \( GH \)) and the radii \( DF \) and \( DG \) of the circle at points of tangency to lines \( FH \) and \( GH \).
2. Since the angles at \( D \) are supplementary:
\[
\angle FDG + \angle FHG = 180^\circ
\]
3. Therefore, substituting the value we have:
\[
136^\circ + \angle FHG = 180^\circ
\]
4. We can subtract \( 136^\circ \) from \( 180^\circ \):
\[
\angle FHG = 180^\circ - 136^\circ
\]
\[
\angle FHG = 44^\circ
\]
Thus, the measure of \( \angle FHG \) is \( \boxed{44^\circ} \).