In a given triangle, angle bisectors AK and BM are drawn so that AK = BM = AB. Find the measure of each angle of the triangle.

36°, 72°, 72°

Step-by-step explanation:

let ∠BAC=A   ∠ABC=B   ∠ACB=C

AB = BM (known)    ∠BAM = ∠AMB = A

AB = AK                   ∠ABK = ∠AKB = B

A = B/2 + C  (exterior angle, ΔBMC)   ...(1)

B = A/2 + C  (exterior angle, ΔAKC)    ...(2)

A - B = (B - A) / 2     (1) - (2)

2A - 2B = B - A          3A = 3B          A = B

A+B+C = 180              2A + C = 180 .....(3)

(180-A) + B/2 + C = 180     (ΔBMC, sum of interior angle)

-A + B/2 + C = 0        -A/2 + C = 0   .....(4)

(3) - (4):   5/2 A = 180          A = 2*180/5 = 72

B = 72        C = 180-72-72 = 36

sorry not sure

step-by-step explanation: