As shown in the figure, the quadrilateral ABCD is a square, AB is an equilateral triangle with an edge to the square, Abe, CE and DB intersect at point F, then ∠ AFD=______ Degree

As shown in the figure, the quadrilateral ABCD is a square, AB is an equilateral triangle with an edge to the square, Abe, CE and DB intersect at point F, then ∠ AFD=______ Degree

In △ CBF and △ ABF, BF = BF ∠ CBF = BC = baabf, ≌ CBF ≌ ABF (SAS), ≌ BAF = BCE = 15 ° and ≌ ABF = 45 ° and ≌ AFD is the outer angle of △ AFB Therefore, the answer is 60

As shown in the figure, take two points E and F on the sides DC and BC of rectangular ABCD, so that AE bisects ∠ DAF. If ∠ BAF = 60 °, calculate the degree of ∠ DAE?

In rectangular ABCD, ∠ bad = 90 degree
∵∠BAF=60°
∴∠DAF=90°-∠BAF=30°
AE bisection of DAF
Therefore, DAE = DAF = 15 degree

Fold the rectangle ABCD along AE so that point d falls at point F on the edge of point BC. If BC = 8cm, BAF = 40 degrees ∠ DAF = 40 degrees, calculate the degree of ∠ DAE and the length of AF

It can be seen from the folding that: ∠ DAE = ∠ FAE = 1 / 2 ∠ DAF = 1 / 2 (90 ° - ∠ BAF) = 25 ° (∠ DAF redundant),
AF=AD=BC=8㎝.

As shown in the figure, in rectangular ABCD, △ Abe is folded along AE, so that point B falls at point F on DC side. If AB = 8cm, ∠ DAF = 48 °, calculate the degree of ∠ BAE and the degree of AF

∠BAE=21°,AF=8cm

As shown in the figure, in the isosceles trapezoid ABCD, ad ∥ BC, ad = 2cm, BC = 8cm, e is the midpoint of the lumbar CD, be divides the trapezoid into two parts, and the difference of its perimeter is 3cm
(1) Find the ratio of abed and △ BEC of the waist AB length (2) quadrilateral

Through D for DF ⊥ BC, BC and F, through e for eg ⊥ BC, BC and GDC ⊥ 178; = DF ⊥ 178; + FC ⊥ 178; DC = AB > 3C (abed) = AD + de + be + AB = AD + (1 / 2) CD + be + AB = AD + (3 / 2) AB + bec (BEC) = be + CE + BC = be + (1 / 2) CD + BC = be + (1 / 2) AB + Bcc (abed) - C (BEC) = (ad-bc) + AB if C (abed) - C (

As shown in the figure, in the trapezoidal ABCD, ad ∥ BC, ab = DC, ∠ ABC = 80 ° e is a point on the lumbar CD, connecting be, AC and AE. If ∥ ACB = 60 ° and ∥ EBC = 50 °, calculate the degree of ∥ EAC

In ladder ABCD, ad ∥ BC, ab = DC, ∥ AC = BD, ∥ DCB = ∠ ABC = 80 °, ∥ ACB = 60 °, ∥ BCF, △ ADF are equilateral triangles, ∥ ACD = ∠ DCB - ∥ ACB = 80 ° - 60 ° = 20 °, ∥ BEC = 180 ° - CBE - ∥ DCB = 180 ° - 50 ° - 80 ° = 50 ° = CBE, ∥ CB = CE = CF, ∥ e, F and B are on the circle with C as the center and CE as the radius. Take any point m at ⊙ C, ∵∠ DCB = 80 °, ∵∠ M = 12 ∠ BCD = 40 °, DFE = ∠ M = 40 ° (any external angle of the inscribed quadrilateral in the circle is equal to its internal diagonal), ∵∠ CDB = 180 ° - ∠ DBC - ∠ DCB = 180 ° - 60 ° - 80 ° = 40 °,