Final answer:
In triangle ABC, let D and E be the points on sides AB and AC such that AD = DB and AE = EC. Let P be the midpoint of segment DE. To find the degree measure of the acute angle formed by lines BP and CP, we can consider the similar triangles BPD and CPD. The acute angle formed by lines BP and CP is 90 degrees.
Step-by-step explanation:
In triangle ABC, let D and E be the points on sides AB and AC such that AD = DB and AE = EC. Let P be the midpoint of segment DE. To find the degree measure of the acute angle formed by lines BP and CP, we can consider the similar triangles BPD and CPD. Since BP and CP are medians of these triangles, they divide the opposite side DE in the ratio 2:1. Therefore, the measure of angle BPC is equal to half the measure of the angle BPD.
Since AD = DB, triangle ABD is isosceles and angle ABD is equal to angle BAD. Similarly, since AE = EC, triangle ACE is isosceles and angle ACE is equal to angle CAE. Therefore, angle BPD is equal to angle BAP + angle DAB = angle CAP + angle CAE. Since BPD and CPD are similar triangles, angle BPD is twice the measure of angle BPC, and angle BPC is twice the measure of angle BPD. Therefore, angle BPD = angle BPC = angle BAP + angle DAB = angle CAP + angle CAE.
Since angle BAD = angle CAE and angle BAP = angle CAP, we have angle BPD = angle BPC = angle BAD + angle BAP. Since the sum of the angles in a triangle is 180 degrees, angle BAD + angle BAP + angle BPC = 180 degrees. Since angle BPD = angle BPC, we have angle BAD + angle BAP + angle BPD = 180 degrees. Since angle BPD = angle BPC = angle BAD + angle BAP, we can substitute this into the equation to get angle BAD + angle BAP + angle BAD + angle BAP = 180 degrees. Simplifying, we find that 2(angle BAD + angle BAP) = 180 degrees, or angle BAD + angle BAP = 90 degrees.
Therefore, the acute angle formed by lines BP and CP is 90 degrees, and the correct answer is
B) 90 degrees.