practice problems pending

Q1. Let X be the midpoint of the side AB of △ABC. Let Y be the midpoint of CX. Let BY cut AC at Z. Prove that AZ=2ZC.

Q2. ABC is a equilateral triangle with vertex A fixed and B moving in a given straight line. Find the locus of C.

Solution 1:
Solution A:
Mass point geometry.
Let weight of C Be 2, since Y is m.p. of CX, weight of X is also 2.
X is m.p. of AB so weight of A,B is 1 each.
Now weight of A is 1, C is 2 so AZ = 2CZ, H.P.

Solution B:
Draw a line from X parallel to BZ. Let it cut AC at W.

In triangle ABZ, W is midpoint of AZ because X is mp of AB and XW || BZ.
IN triangle CXW, YZ || XW => Z is m.p. of CW. H.P.

Solution C:
Menelaus's theorem:
BYZ is a transversal for triangle AXC, and B sits on AX extended.
AB/BX * XY/YC * CZ/ZA = 1 = 2 * 1 * CZ/ZA => ZA = 2CZ H.P.

Solution 2:
B moves in a fixed line and AC is nothing but AB rotated by 60 degree clockwise or counterclockwise.
So locus of C will be 2 lines which make 60 degree angle with B's locus line in 2 directions.