Triangle similarity question

Question

I've been trying to solve this question for like 40 mins straight and can't seem to get anywhere. I tried drawing a parallel to |KM| from C to |AB| but that didn't seem to help. I just can't see a way to get to |MN|.

Answer 1

Using Menelaus' theorem, (AB)(KN)(MC)/(BK)(NM)(CA) = 1 (using unsigned lengths) Similarly, (AM)(CN)(BK)/(MC)(NB)(KA) = 1.

Using the second equation, AM/MC = 6 and hence AC/MC = 5. Substituting into the first equation, we find KN/NM = 5/3. Together with MK = 24, we get MN = 9.

Answer 2

Construction: (1) Join BM; (2) Let T be the midpoint of KA. Join TN; (3) Join AN.

From the construction, BK = KT = TA; and therefore, all yellow shaded (including light, normal and dirty) triagnles are equal in area.

In ⊿s ABN and ACN, because BN : NC = 3 : 1, all yellow shaded (including light, normal ,dirty AND orange) triangles are equal in area (= Y say).

In ⊿s MBN and MCN, because BN : NC = 3 : 1, area of the Red-shaded : area of the Pink-shaded = 3 : 1. We will denote this result as $\frac {R}{P} = \frac {3}{1}$……….(1)

Similarly, in ⊿s MBK and MAK, we have $\frac {Y + R}{3Y + P} = \frac {1}{2}$……….(1)

Eliminating P from (1) and (2), we have $\frac {Y}{R} = \frac {5}{3}$

This means $\frac {KN}{NM} = \frac {5}{3}$

Therefore, $NM = \frac {3}{5 + 3} × 24 = 9$