Explanation:
good job then !
the first thing I see in the graphic is that
BG is half of BD (as a major definition of diagonals in a parallelogram).
BD on the other hand is the Hypotenuse (the side opposite of the 90° angle at A) of the triangle ABD.
to be able to use Pythagoras to get BD, I need to find the length of AB.
this I get from the triangle ABE. this is also a right-angled triangle. AE = 4cm. the angle B1 = 50°. the angle A = 90°.
so, the angle E1 = 180 - 90 - 50 = 40° (remember, the sum of all angles in a triangle is always 180°).
now we use the law of sine :
a/sin(A) = b/sin(B) = c/sin(C)
with a, b, c being the sides of the triangle, and A, B, C being the corresponding opposite angles.
so, we have here
AE / sin(50) = AB / sin(40)
4 / sin(50) = AB / sin(40)
AB = 4×sin(40) / sin(50) = 3.356398525... cm
so (Pythagoras),
BD² = AB² + AD² = 3.356398525...² + 9² =
= 11.26541106... + 81 = 92.26541106...
BD = 9.60548859... cm
BG = BD/2 = 4.802744295... cm
and now look at that.
because of the fact that I did not have to go through the other points, I chose the first approach that went through my mind.
and it showed that your teacher created an impossible (contradicting) scenario :
the angle of B1 does not fit to the ratios of the various line segments.
as you correctly calculated CD as 3 cm, that must mean that AB is also 3 cm.
that makes
BD² = 9² + 3² = 81 + 9 = 90
BD = 9.486832981... cm
BG = BD/2 = 4.74341649... cm
PLEASE, USE THIS VALUE.
there is a difference to my previous calculation. and the reason for this is : the 50° for B1 are not correct.
with AE = 4, AB = 3 and then BE = 5, we get
AE / sin(B1) = BE / sin(A)
4 / sin(B1) = 5 / sin(90) = 5
sin(B1) = 4/5 = 0.8
B1 = 53.13010235...°
and not 50° !!!!