SOLUTION: 1.M is the mid-point of a line segment AB;AXB and MYB are equilateral triangles on opposite sides of AB;XY cuts AB at Z.Prove that:AZ=2 ZB. 2.In trapezium ABCD,AB//DC and DC=2 AB.

Algebra ->  Geometry-proofs -> SOLUTION: 1.M is the mid-point of a line segment AB;AXB and MYB are equilateral triangles on opposite sides of AB;XY cuts AB at Z.Prove that:AZ=2 ZB. 2.In trapezium ABCD,AB//DC and DC=2 AB.      Log On


   

Question 550587: 1.M is the mid-point of a line segment AB;AXB and MYB are equilateral triangles on opposite sides of AB;XY cuts AB at Z.Prove that:AZ=2 ZB.
2.In trapezium ABCD,AB//DC and DC=2 AB.EF,drawn parallel to AB cuts AD in F and BC in E such that 4 BE=3 EC.Diagonal DB intersects FE at point G.Prove that:7 EF=10 AB.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
1. Since M is the midpoint of AB, AM=MB and AB=2AM=2MB
Since ABX and ABY are equilateral, AB=BX=AX=2MB=2BY=2MY
Triangles BXZ and BZY have congruent vertical angles at Z, and congruent 60 degree angles ZBX and ZMY. They are similar, and since BX=2BY, ZB=2MZ.
Then AM=MB=MZ+ZB=3MZ.
So AZ=AM+MZ=3MZ+MZ=4MZ,
and since ZB=2MZ, AZ/ZB=2MZ/2MZ=2
AZ/ZB=2 --> AZ=2ZB

2. I used to call ABCD a trapezium too, but I'm going to call it a trapezoid, because I live in the USA now.
The red line FE divides each of the 3 slanted lines into two segments with lengths in the ratio 4:3.
On line BC
4+BE=3+EC --> BE=%283%2F4%29EC BC=BE%2BEC=%287%2F4%29EC --> EC=%284%2F7%29BC --> BE=%283%2F7%29BC
Similar ratios can be written for the segments on lines DB and AD.
Diagonal DB divides the trapezoid into 2 triangles (BCD and ABD). Each of those triangles has a smaller similar triangle to one side of red line FE (BEG and FGD respectively).
The ratio of the side lengths of BEG and BCD is 3:7, so GE=%283%2F7%29DC.
The ratio of the side lengths of FGD and ABD is 4:7, so FG=%284%2F7%29AB.
Since DC=2 AB,

FE=%2810%2F7%29AB --> 7FE=10AB