Triangle ABC, angle A = 37 degrees, angle C = 65 degrees. A line MN is drawn through vertex B

Triangle ABC, angle A = 37 degrees, angle C = 65 degrees. A line MN is drawn through vertex B and is parallel to AC. Find the angle MBD, where BD is the bisector of angle ABC.

Knowing that the sum of all the angles of the triangle is 180 °, we can find the unknown angle – ∠В.
180 ° – ∠А – ∠С = 180 ° – 37 ° – 65 ° = 78 °.
By hypothesis, ∠В has a bisector BD.
By the properties of the bisector, we know that it divides the angle exactly in half.
Hence, we can find the value of ∠ABD.
78 °: 2 = 39 °.
It remains to find the angle ∠МВА, for this we consider the triangle МВА.
Since by condition the line MV is parallel to AD, we can conclude that ∠АМВ = ∠MAD = 90 °.
We can find ∠MAV:
90 ° – 37 ° = 53 °.
Find the third corner:
∠MBA = 180 ° – ∠AMB – ∠MAB = 180 ° – 90 ° – 53 ° = 37 °.
Now we can find the required ∠MBD:
∠MBD = ∠ABD + ∠MBA = 39 ° + 37 ° = 76 °.
Answer: ∠MBD = 76 °.