Question 550587
1.{{{drawing(200,300,-2.4,2.4,-2.6,4.6,
triangle(-2,0,2,0,0,3.464),
triangle(0,0,2,0,1,-1.732),
red(line(0,3.464,1,-1.732)),
locate(-0.1,3.8,X), locate(0.9,-1.7,Y),
locate(-2.1,0,A), locate(2.1,0.3,B),
locate(-0.15,0,M), locate(0.65,0.4,Z)
)}}} 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.
{{{drawing(300,300,-1,9,-1,9,
grid(1),
blue(line(0,8,8,8)), blue(line(1,1,5,1)),
blue(line(0,8,1,1)), blue(line(8,8,5,1)),
green(line(0,8,5,1)), red(line(4/7,4,6&2/7,4)),
locate(0.7,1,A),locate(5.1,1,B),locate(8.2,8.5,C),locate(0.2,8.5,D),
locate(0.2,4,F),locate(3,4.5,G),locate(6.5,4,E)
)}}} 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=(3/4)EC}}} {{{BC=BE+EC=(7/4)EC}}} --> {{{EC=(4/7)BC}}} --> {{{BE=(3/7)BC}}}
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=(3/7)DC}}}.
The ratio of the side lengths of FGD and ABD is 4:7, so {{{FG=(4/7)AB}}}.
Since DC=2 AB,
{{{FE=FG+GE=(4/7)AB+(3/7)DC=(4/7)AB+(3/7)(2*AB)=(4/7)AB+(6/7)AB=(10/7)AB}}}
{{{FE=(10/7)AB}}} --> {{{7FE=10AB}}}