m∠BAC = 27°
Solution:
ABCD is a quadrilateral.
AB and CD are parallel lines.
Given m∠BCD = 54°
AC bisect ∠BCD.
m∠DCA + m∠CAB = m∠BCD
m∠DCA + m∠DCA = 54° (since ∠ACB = ∠DCA)
2 m∠DCA = 54°
Divide by 2 on both sides, we get
m∠DCA = 27°
AB and CD are parallel lines and AC is the transversal.
If two parallel lines cut by a transversal, then the alternate interior angles are equal.
m∠BAC = m∠DCA
m∠BAC = 27°
Hence m∠BAC = 27°.
