Proof of Mathematics and Geometry in Grade Two (Attached)

Proof: For convenience, let ∠BAD=∠ 1, ∠ACF=∠2, ∠DEB=∠3, ∠EAB=∠4, ∠DCG =∞.

Because: BD = AF, AB = AC, ∠ Abd = ∠ CAF = 60.

So: triangle ABD and triangle CAF are congruent.

So: ∠ 1=∠2, and FC=AD.

Because: Abd = AED = 60.

So: AEBD is a four-point circle.

So: ∠ 1=∠3

So there are: ∠ 1=∠2=∠3.

From the * * * circle: ∠10 = ∠1= ∠ Abd = ∠ FAC = 60.

Therefore: ∠ 7 = 60+∠ 3, ∠ 6 = 60+∠ 1 and ∠ 8 = 60+∠ 2.

So: ED parallel FC is obtained by ∠7=∠8.

Because FC=AD=ED

So: Quadrilateral EDCF is a parallelogram. (A set of quadrilaterals with parallel and equal opposite sides is a parallelogram)