Question

In isosceles △ABC with AC = AB, points D ∈ AB and E ∈ AC so that DB = EC. Point P is the intersection point of DC and EB, m∠ADC = 107°, m∠ABE = 45°. Find m∠BAP.

Please show work.

Answer 1

9514 1404 393

Answer:

14°

Explanation:

∠ADP is an external angle to ΔDBP, so ...

∠ADP = ∠DBP + ∠DPB

107° = 45° + ∠DPB

∠DPB = 107° -45° = 62°

∠DPB is an external angle to isosceles triangle PBC, so ...

∠DPB = 2×∠PBC

62° = 2×∠PBC

∠PBC = 62°/2 = 31°

Of course, the angle sum theorem applies at vertex B, so ...

∠ABC = ∠ABP +∠PBC

∠ABC = 45° +31° = 76°

AP is the angle bisector of angle A, so, if extended to BC would be perpendicular to BC. Then ∠BAP is complementary to ∠ABC:

∠BAP = 90° -∠ABC

∠BAP = 90° -76°

∠BAP = 14°