The quadrilateral ABCD is the inner boundary quadrilateral of the circle, the diagonal AC and BD intersect at e, the extended DA and CB intersect at F, the angle CAD = 60 degrees, DC = De, it is proved that a is the outer center of the triangle bef How does angle F + angle AEB equal to 60 degrees?

The quadrilateral ABCD is the inner boundary quadrilateral of the circle, the diagonal AC and BD intersect at e, the extended DA and CB intersect at F, the angle CAD = 60 degrees, DC = De, it is proved that a is the outer center of the triangle bef How does angle F + angle AEB equal to 60 degrees?

DC = De, angular DCE = angular Dec, that is, angular AEB = angular Abe, so AE = ab
Angle F + angle AEB = 120 °, angle FBA + angle Abe = 180 ° - angle CBD = 180 ° - angle CDA = 120 °
That is angle FBA + angle AEB = 120 degrees
So angle f = angle FBA
That is ab = AF = AE, a is the outer center of the triangle bef
Sorry, I didn't see it

As shown in the figure, in the quadrilateral ABCD, ad = CB, ∠ ACB = CAD

Proof 1: in △ ABC and △ CDA, ∵ ad = BC, ∵ ACB = ∵ CAD, AC = AC, ≌ ABC ≌ CDA & nbsp; & nbsp; (SAS).. AB = CD. Proof 2: ∵ ACB = ∵ CAD, ∥ ad ∥ BC. ∵ ad = BC, ≌ quadrilateral ABCD is parallelogram. ≌ AB = CD

It is known that in a quadrilateral ABCD, AC and BD intersect point O, and the extension lines of Da, BC and DC intersect at points F, e and G through O. to prove: go & sup2; = Ge · GF
(hint: the title of the test paper is: parallel line is proportional to line segment + I don't know what the similarity is.)

A + B = B + a add exchange rate! Use this set ~!

The following quadrilateral ABCD is a square, the triangle CEF is 6 cm larger than the triangle ADF, ab = 6 cm, find the length of CE

For example: (as shown in the figure) the side length of square ABCD is 5cm, and the area of triangle CEF is 5 square centimeters larger than triangle ADF. How many centimeters is CE? Because triangle ADF is a part of square ABCD, and triangle CEF is a part of big triangle Abe. So "the area of triangle CEF is 6 square centimeters larger than triangle ADF

In the parallelogram ABCD, when e is a point on AB, f is a point on CD, and AE = CF, as shown in the figure, we prove that AF = CE, ∠ EAF = ∠ EAF
Help

In the parallelogram ABCD, ab ‖ CD,
Because AE = CF,
So the quadrilateral aecf is a parallelogram,
So AF = CE, ∠ EAF = ∠ ECF

As shown in the figure, the side length of square ABCD is 4, which is the midpoint of BC side, f is the point on DC side, and DF = 14dc, AE and BF intersect at G point. Find the area of △ ABG

As shown in the figure, GM ⊥ AB is made at point m through point G, let GM = x, because Tan α = & nbsp; BEAB = gmam = 24, so am = 2x, and Tan β = bccf = gmbm = 43, so BM = 3x4, am + BM = AB = 4, that is, 2x + 3x4 = 4, the solution is x = 1611. So s △ ABG = 12 × ab × GM = 12 × 4 × 1611 = 3211

As shown in the figure, the side length of square ABCD is 4, which is the midpoint of BC side, f is the point on DC side, and DF = 14dc, AE and BF intersect at G point. Find the area of △ ABG

As shown in the figure, GM ⊥ AB is made at point m through point G, let GM = x, because Tan α = & nbsp; BEAB = gmam = 24, so am = 2x, and Tan β = bccf = gmbm = 43, so BM = 3x4, am + BM = AB = 4, that is, 2x + 3x4 = 4, the solution is x = 1611. So s △ ABG = 12 × ab × GM = 12 × 4 × 1611 = 3211