In the diagram below, start overline, A, D, end overline, \parallel, start overline, E, G, end overline, comma

Question

AD
∥
EG
, start overline, B, H, end overline, is perpendicular to, start overline, F, C, end overline, comma
BH
⊥
FC
, and m, angle, A, B, H, equals, 147, degrees, .m∠ABH=147
∘
. Find m, angle, E, F, H, .m∠EFH.

Answer 1

Given the information, we have:

1. \( AD \parallel EG \) (lines \( AD \) and \( EG \) are parallel).
2. \( BH \perp FC \) (line \( BH \) is perpendicular to line \( FC \)).
3. \( m \angle ABH = 147^\circ \).

To find \( m \angle EFH \), we start by analyzing the angles involved.

Since \( BH \) is perpendicular to \( FC \), we know that:

\[ m \angle ABH + m \angle BHF = 90^\circ. \]

Thus, we can calculate \( m \angle BHF \):

\[ m \angle BHF = 90^\circ - m \angle ABH = 90^\circ - 147^\circ = -57^\circ. \]

Since angles cannot be negative, it actually indicates a relationship based on linear pairs or angles being measured in the opposite direction.

Now, we should notice that \( m \angle ABH = 147^\circ \) and using the fact that angles formed by a transversal intersecting parallel lines are congruent or supplementary:

Considering \( AD \parallel EG \) and the transversal \( BH \), we have:

\[ m \angle ABH + m \angle EFH = 180^\circ. \]

Since \( m \angle ABH = 147^\circ \):

\[ m \angle EFH = 180^\circ - 147^\circ = 33^\circ. \]

Therefore, the measure of angle \( EFG \) is:

\[ \boxed{33^\circ}. \]