# 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.