Median BM and bisector AP of triangle ABC intersect at point K, side length AC refers to the length of side AB as 9: 7

Median BM and bisector AP of triangle ABC intersect at point K, side length AC refers to the length of side AB as 9: 7. Find the ratio of the area of the triangle ABK to the area of the quadrangle KPCM

BM is the median of the triangle ABC, therefore, it divides this triangle into two equal in area triangle (median property).
SABM = SCMB = SABC / 2
Consider the triangle ABM.
SABK + SAKM = SABM = SABC / 2
AP is a bisector, by the bisector theorem we can write AM / AB = KM / BK.
By the condition of the problem, AC / AB = 9/7, therefore, 2 AM/AB=9/7 => AM / AB = 9/14 => KM / BK = 9/14
Because the area of ​​the triangle is calculated by the formula S = 1/2 * h * a, where a is the base and h is the height, we can write:
SAKM = 1/2 * h * KM = 1/2 * h * ((9/14) * BK) = 9/14 * (1/2 * h * BK) = 9/14 * SABK (since height h for these triangles is common)
SABK + SAKM = SABM = SABC / 2
SABK + 9/14 * SABK = SABC / 2
23/14 * SABK = SABC / 2
SABK = 14SABC / 46
By the same bisector property for triangle ABC, we obtain that AC / AB = CP / PB
AC / AB = 9/7 (by the condition of the problem) => CP / PB = 9/7 therefore, CP = 9 * PB / 7
SAPC = 1/2 * h * PC = 1/2 * h * (9 * PB / 7) = 9/7 * (1/2 * h * PB) = 9/7 * SABP,
SABP + SAPC = SABC
SABP + 9/7 * SABP = SABC
16/7 * SABP = SABC
SABP = 7/16 * SABC
Next, find the area of ​​the triangle BPK:
SBPK = SABP-SABK
We previously found that SABK = 14SABC / 23
SBPK = 7SABC / 16-14SABC / 46 = 322SABC / 736-224SABC / 736 = 98SABC / 736 = 49SABC / 368
Find the area of ​​the quadrangle KPCM:
SKPCM = SCMB-SBKP
SKPCM = SABC / 2-49SABC / 368, (we found the CMB area earlier),
SKPCM = 184SABC / 368-49SABC / 368 = 135SABC / 368
The ratio of the areas of ABK to KPCM = (14SABC / 46) / (135SABC / 368) = (14 * 368) / (46 * 135) = (14 * 8) / 135 = 112/135
Answer: the ratio of the area of ​​the triangle ABK to the area of ​​the quadrangle KPCM = 112/135.