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In order to know that the sum of the inner angles of the hexagon ABCDEF is equal, please judge the size of AB plus BC and de plus EF
If we extend BC. De, FA to intersect P. Q, R. respectively, then △ PQR is an equilateral triangle (because every inner angle of a hexagon is equal, it is 120 ° and every outer angle is 60 °, ﹤ EFR. △ DCQ, △ ABP are equilateral triangles
As shown in the figure, in the hexagon ABCDEF, it is proved that BC / / Fe
Because the sum of the angles in a hexagon is 720 degrees
So ∠ a + ∠ B + ∠ C + ∠ D + ∠ e + ∠ f = 720 °
Because ∠ a = ∠ D ∠ B = ∠ e ∠ C = ∠ f
Therefore, 2 ∠ C + 2 ∠ D + 2 ∠ e = 720 °
∠E+∠C+∠D=360°
Connect CE two points
Because ∠ D + ∠ ECD + ∠ CED = 180 °
So ∠ BCE + ∠ FEC = 180 °
So BC / / Fe
As shown in the figure, in the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, CM bisection ∠ BCD intersects AF in M, FN bisection ∠ AFE intersects CD in N. try to judge the position relationship between cm and FN, and explain the reasons
CM∥FN．
Let ∠ a = ∠ d = α, ∠ B = ∠ e = β, ∠ BCM, ∠ AMC, ∠ 3, ∠ AFN, ∠ 2,
∵ the sum of internal angles of hexagon is 720 °,
∴2∠1+2∠2+2α+2β=720°，
∴∠1+∠2=360°-α-β，
And in the quadrilateral ABCM, ∠ 1 + ∠ 3 = 360 ° - α - β,
∴∠2=∠3，
∴CM∥FN．
As shown in the figure, in the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, CM bisection ∠ BCD intersects AF in M, FN bisection ∠ AFE intersects CD in N. try to judge the position relationship between cm and FN, and explain the reasons
CM∥FN．
Let ∠ a = ∠ d = α, ∠ B = ∠ e = β, ∠ BCM, ∠ AMC, ∠ 3, ∠ AFN, ∠ 2,
∵ the sum of internal angles of hexagon is 720 °,
∴2∠1+2∠2+2α+2β=720°，
∴∠1+∠2=360°-α-β，
And in the quadrilateral ABCM, ∠ 1 + ∠ 3 = 360 ° - α - β,
∴∠2=∠3，
∴CM∥FN．
As shown in the figure, in the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, CM bisection ∠ BCD intersects AF in M, FN bisection ∠ AFE intersects CD in N. try to judge the position relationship between cm and FN, and explain the reasons
CM∥FN．
Let ∠ a = ∠ d = α, ∠ B = ∠ e = β, ∠ BCM, ∠ AMC, ∠ 3, ∠ AFN, ∠ 2,
∵ the sum of internal angles of hexagon is 720 °,
∴2∠1+2∠2+2α+2β=720°，
∴∠1+∠2=360°-α-β，
And in the quadrilateral ABCM, ∠ 1 + ∠ 3 = 360 ° - α - β,
∴∠2=∠3，
∴CM∥FN．
In the hexagon ABCDEF, angle a = angle D, angle B = angle e, CM bisection angle BCD intersects AF at point m, FN bisection angle AFE intersects CD at point n. try to judge the position relationship between cm and FN, and explain the reasons. There is also a picture... Which is the last question of triangle test volume 2 (7.7.4) of the first grade education edition of junior high school
From angle a = angle D, angle B = angle e, we know ∠ P = ∠ Q, extend EF and cm to R, extend BC and FN to s, △ CPM and △ RCQ, ∠ PCM = ∠ fcq (angular bisector), ∠ P = ∠ Q, so ∠ PMC = ∠ CRQ, and ∠ PMC = ∠ RMF (opposite top angle), so ∠ PFQ = ∠ RMF +
As shown in the figure, in the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, CM bisection ∠ BCD intersects AF in M, FN bisection ∠ AFE intersects CD in N. try to judge the position relationship between cm and FN, and explain the reasons
CM∥FN．
Let ∠ a = ∠ d = α, ∠ B = ∠ e = β, ∠ BCM, ∠ AMC, ∠ 3, ∠ AFN, ∠ 2,
∵ the sum of internal angles of hexagon is 720 °,
∴2∠1+2∠2+2α+2β=720°，
∴∠1+∠2=360°-α-β，
And in the quadrilateral ABCM, ∠ 1 + ∠ 3 = 360 ° - α - β,
∴∠2=∠3，
∴CM∥FN．
As shown in the figure, in the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, try to explain that AF is parallel to CD Make use of junior high school knowledge to seek, had better be simple
Connect FC, in quadrilateral abcf and quadrilateral DEFC, ∠ a + ∠ B + ∠ 1 + ∠ 4 = ∠ D + ∠ e + ∠ 2 + ∠ 3 = 360
According to the title: ∠ a = ∠ D ∠ B = ∠ e, so ∠ 1 + ∠ 4 = ∠ D + ∠ e + ∠ 2 + ∠ 3
And ∠ C = ∠ F, so ∠ 1 + ∠ 2 = ∠ 3 + ∠ 4
The solution is: ∠ 1 = ∠ 3
So AF is parallel to CD
As shown in the figure, in hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to explain that AF and CD are parallel
Proof: connect ad
∵ the sum of inner angles of quadrilateral is 360 ゜
∴∠2+∠3=∠1+∠4．
And ? 2 + ∠ 4 = ∠ 1 + ∠ 3
∴∠1=∠2
∴AF∥CD．
As shown in the figure, in the hexagon ABCDEF, AF ∥ CD, ab ∥ ed, ∠ a = 140 °, B = 100 °, e = 90 °. Calculate the degrees of ∠ C, ∠ D and ∠ F
Through point BG ∥ AF, through point C as CH ≓ AB, ∵ AF ∥ CD, ab ∥ ed, ? BG ∥ AF ∥ CD, CH ∥ ab ∥ De, ? a + ∥ ABG = 180 °, BCD + CBG = 180 °, that is, ? ABC + ∥ BCD = 360 °, a = 140 °, ABC = 100 °, BCD = 120 °, similarly, ∠ ABC + ∠ B
RELATED INFORMATIONS
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