Finding m<HFG:
Look at triangle HFG.
It has angles HFG, FGH, and GHF.
The sum of the measure of the angles of a triangle equals 180 deg.
m<FGH = 5x + 4
<GHF is a right angle, so it measures 90 deg.
m<HFG + m<FGH + m<GHF = 180
m<HFG + 5x + 4 + 90 = 180
m<HFG + 5x + 94 = 180
m<HFG = 180 - 94 - 5x
m<HFG = 86 - 5x
m<HFG = (86 - 5x)°
Finding m<BCF:
Quadrilateral IBCF is a rectangle because of the right angles shown.
BC is parallel to IH.
<A is congruent to <CBF because of corresponding angles.
m<A = m<CBF = 10x + 3
For triangle AGC,
m<A + m<G + m<C = 180
10x + 3 + 5x + 4 + m<C = 180
m<C + 15x + 7 = 180
m<C = 173 - 15x
m<BCF = (173 - 15x)°
Finding angle FIB:
BF and IH are parallel.
m<BFI = m<FIH = 4x + 3 by alternate interior angles.
Now look at triangle FBI.
m<IFB + m<FBI + m<FIB = 180
4x + 3 + 90 + m<FIB = 180
m<FIB + 4x + 93 = 180
m<FIB = 87 - 4x
m<FIB = (87 - 4x)°
Now look at <EFG. <EFG is an exterior angle of triangle FGI.
Theorem: The measure of an exterior angle of a triangle equals the sum of the measures of the remote interior angles.
For exterior <EFG, the remote interior angles are <FIG and <FGI.
According to the theorem above,
m<FIG + m<FGI = m<EFG
Now we substitute the measure of each angle and solve for x.
4x + 3 + 5x + 4 = 9x + 7
