practice problems pending

Q1. While applying Menelaus on a triangle ABC, can the traversal pass through one of its vertices?
S1. No. In one of the fractions numerator would become zero. In another denominator.

Q2. The points (X, Y) are taken on (CA, AB) respectively of triangle ABC. If (BX) and (CY) meet at (P) and AX/CX = BY/AY = 1/2 then find BP/PX.
S2. Triangle XAB, Traversal YPC. Apply Menelaus. Answer: 3/4

Q3. In △ABC the points E, F, G are on AB, BC & CA respectively, such that AE/EB = BF/FC = CG/GA = 1/3. The (K, L, M) are the intersections of (AF) & (CE); (BG) & (AF); (CE) & (BG) respectively.

If ([ABC] = 1), find ([KLM] = ?)
S3.
Here we need to use mass points on 2 Cevians at a time. We can't apply on all 3 together since they are not concurrent.
Our aim is to find [KLM] like this: 
[KLM] = [ABC] - [AKC] - [ALB] - [BMC]
For [AKC], use mass points on cevians AF,CE to get the ratio AK:KF.
Then [CKA]:[CKF] = AK:KF
But [CKA] + [CKF] = [AFC]
And [AFC]/[ABC] = FC/BC
So [AFC] = 3/4
So we can get [CKA] = 3/13.

Similarly get [ALB] = 3/13 = [BMC] since all the ratios are same.
So [KLM] = 1-9/13 = 4/13 = Answer

You can also use Routh's theorem which is directly applicable for this very problem.

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