As shown in the figure, in the parallelogram ABCD, e and F are two points on the diagonal BD, and be = DF, connecting AE, AF, CE and cf. what kind of quadrilateral is the quadrilateral aecf, which shows your truth

As shown in the figure, in the parallelogram ABCD, e and F are two points on the diagonal BD, and be = DF, connecting AE, AF, CE and cf. what kind of quadrilateral is the quadrilateral aecf, which shows your truth

∵ quadrilateral ABCD is a parallelogram, ∵ ab ‖ CD, ab = CD, ∵ Abe = ∠ CDF, ∵ be = DF, ≌ Abe ≌ CDF, ≌ AE = CF, similarly: CE = AF, ≌ quadrilateral aecf is a parallelogram

As shown in the figure, the quadrilateral ABCD is a parallelogram, be ‖ DF, and intersects diagonals AC at points E and f respectively, connecting ed and BF

It is proved that the ∵ quadrilateral ABCD is a parallelogram, ∵ AB = CD, ab ∥ CD, ∵ be ∥ DF, ∵ bef = ∥ EFD, ∵ bef + ∥ AEB = 180 °, ∥ EFD + ∥ DFC = 180 °, ∥ AEB = ∥ CFD. ≌ Abe ≌ CDF (AAS) ∥ be = DF. ∥ bfde is a parallelogram. ∥ de ∥ BF. ∥ 1 = ∥ 2

Parallelogram ABCD, be ⊥ CD, BF ⊥ ad, e, f are perpendicular feet, ∠ FBE = 60 °, AF = 3cm, CE = 4.5cm, then ∠ C = (), ∠ d = (), ab = (

Parallelogram ABCD, be ⊥ CD, BF ⊥ ad, e, f are perpendicular feet, ∠ FBE = 60 °, AF = 3cm, CE = 4.5cm, then ∠ C = (120 °), ∠ d = (60 °), ab = (6
Be ⊥ CD, then ∠ bed = 90 °
BF ⊥ ad, then ∠ BFD = 90 °
If ∠ FBE = 60 °, then ∠ d = 360 ° - 90 ° - 90 ° - 60 ° = 120 °
∠C＝180°-∠D＝180°-120°＝60°
∠ABF＝90-∠FBE＝90°-60°=30°
AB＝AF/sin30=3/(1/2)=6

In the parallelogram ABCD, e and F are two points on the diagonal BD, AE = CF, and de = BF

Because the quadrilateral ABCD is a parallelogram
So angle Abe = angle CDF, ab = CD
And because AE = CF
So the triangle Abe is equal to the triangle CDF
So be = DF
So bd-df = bd-be
That is BF = De