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