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Known: as shown in the figure, in square ABCD, points E.M and N are on the sides of AB, BC and ad respectively, CE = Mn, ∠ MCE = 35 degrees, find the degree of ∠ anm
Make MF ⊥ ad through M and hand over ad to F,
Because MF = AB = BC
MN=CE
So in RT △ EBC ≌ RT △ NFM
Therefore, anm = FNM = BEC = 90-35 = 55 (degree)
In square ABCD, point E is on ad, point F is on CD, ∠ EBF = 45 °, BG ⊥ EF is on point G
Extend Da to m, make am = CF, connect BM
It is easy to prove △ BCF ≌ △ BAM by SAS
Therefore, CBF = ABM, BF = BM
Because ∠ CBF ＋ Abe = 90 ° - FBE = 90 ° - 45 ° = 45 °
Therefore, MBE = MBA + Abe = CBF + Abe = 45 degree
Therefore, MBE = FBE
So △ BFE ≌ △ BME (SAS)
So ∠ beg = ∠ bea
Because ∠ Bge = ∠ BAE = 90 °
And because be = be
So △ Bge ≌ △ BAE (AAS)
So BG = ab
For reference! Jswyc
In square ABCD, e and F are points on CD and Da respectively, and EF = AF + CE, then ∠ EBF =?
If we extend EC to f 'so that CF' = AF, even BF ', it is easy to prove that two right triangle BAF and BCF' are congruent. So, ∠ ABF = ∠ cbf'bf = bf'be = beef '= EC + CF' = EC + AF = EF, so, △ FBE ≌ f'be, so, ∠ EBF = ∠ EBF ', ∠ EBF = ∠ FBF' / 2, because, ∠ ABF = ∠ CBF ', so, ∠ FBF' = ∠ ABC = 90
As shown in the figure: e and F are the points on the edge CD and Da of square ABCD respectively, and CE + AF = EF, please use the method of rotation to find the size of ∠ EBF
Take △ BCE as the rotation center, rotate 90 ° anticlockwise to make BC fall on the side of Ba, get △ BAM, then ∠ MBE = 90 °, am = CE, BM = be, ∵ CE + AF = EF, ∵ MF = EF, in △ FBM and △ FBE, ≌ be = BMBF = bFEF = MF, ≌ FBM ≌ FBE (S.S.S), ≌ MBF = EBF, ≌ EBF
As shown in the figure, m and N are the midpoint of AD and DC on both sides of the square ABCD, cm and DN intersect at P, and PA = Pb is proved
Certification:
Extend PM and cross the extension line of Ba to point F
Because m and N are the midpoint of AD and CD respectively
Easy to get △ BCN ≌ △ CDM
Then, CBN = MCD
Easy to get ∠ BPF = 90 °
∵ m is the midpoint of AD, BF ∥ CD
∴△MCD≌△MAF
∴CD=AF=AB
｜ PA = AB (the middle line of the hypotenuse of a right triangle is equal to half of the hypotenuse)
As shown in the figure, m and N are the midpoint of AD and DC on both sides of the square ABCD, cm and BN intersect at P
In △ BCN and △ CBM, BC = CD, BCN = cdmcn = DM, ≌ BCN ≌ CDM (SAS), ≌ CBN = DCM, ≌ DCM + BNC = 90 °, CPN = 90 ° & nbsp; and ≌ A is RT △
In square ABCD, M is the point on BC, e is on the extension line of BC, Mn ⊥ am,
In square ABCD, M is a point on BC, e is on the extension line of BC, Mn is perpendicular to am, and the bisector of the intersection angle DCE of Mn is on n
Use the knowledge of grade two in junior high school!!!
Make a point G on AB so that BG = BM
Connect mg
Because the quadrilateral ABCD is a square
So ∠ B is 90 degrees
And BG = BM, so ∠ BGM = ∠ BMG = 45 degrees
Therefore, GAM = 45 degrees
Because NN bisects ∠ DCE, so ∠ DCN = 45 degrees
Therefore, MCN = MCD + DCN = 90 degrees + 45 degrees = 135 degrees
In addition, GAM, NMC and amn complement each other
Therefore, GAM = NMC
So triangle AGM ≌ triangle MNC
So am = Mn
Known: square ABCD, M is the midpoint of AB side, e is a point on the extension line of AB, connect MD, make Mn perpendicular to DM, intersect with angle CBE bisector BN at point n
(1) Verification: DM = Mn
(2) If we change "m is the midpoint of AB" to "m is any point on AB", is "MD = Mn" still valid? Why?
It is proved that in ABCD, M is the midpoint of AB, DF = AF = am = BM ∠ AFM = 45 ° i.e., ∠ DFM = 135bn is the bisector of ∠ CBE ∠ EBN = 45 ° i.e., ∠ MBN = 135 ° so ∠ DFM = ∠ mbnmn is perpendicular to MD ∠ FDM + ∠ amd = 90 ∠ BMN + ∠ amd = 90 ° i.e., ∠ FDM = ∠ BMN and ∠ DFM = ∠ MB
As shown in the figure, the known point F is the midpoint of the side BC of the square ABCD, CG bisects ∠ DCE, GF ⊥ AF
It is proved that: take the midpoint m of AB, connect FM. ∵ point F is the midpoint of the side BC of the square ABCD, ∵ BF = BM, ∵ BMF = 45 °, ∵ AMF = 135 °. ∵ CG bisection ∵ DCE, ∵ GCE = 45 °,
RELATED INFORMATIONS
- 1. The length of square ABCD is 4, M is the moving point on BC, n is the moving point on CD, am is perpendicular to Mn, BM =? ABCN has the largest area
- 2. As shown in the figure, the rectangle ABCD is folded along AE so that point d falls at point F on the edge of BC. If ∠ BAF = 60 °, then ∠ DAE=______ Degree
- 3. As shown in the figure, in diamond ABCD, point E is on BC, and AE = ad, ∠ ead = 2 ∠ BAE, find the degree of ∠ BAE
- 4. It is known that e is the middle point on the side BC of square ABCD, f is a point on CD, AE bisects ∠ BAF
- 5. Known: as shown in the figure, in square ABCD, e is the midpoint of BC, point F is on CD, angle FAE is equal to angle BAE, AF = BC + EC
- 6. The side length of a diamond is 5, and a diagonal line is 6. Now take the straight line where the two diagonal lines are as the coordinate axis to find the four vertices
- 7. In the rectangular trapezoid ABCD, AD / / BC, ∠ a = 90 ° ad = 21cm BC = 24cm E.F. moves from A.C. to D.B. at the same time at 1cm / s and 2cm / s. when they move for how long, is the trapezoid efcd a rectangular trapezoid?
- 8. The area of square ABCD is 120 square centimeters, e is the midpoint of AB, f is the midpoint of BC, what is the area of quadrilateral bghf?
- 9. In trapezoid ABCD, AB / / CD, de / / BC, CD = 7cm, the perimeter of trapezoid is 35cm, and that of triangle ade is 35cm
- 10. It is proved that: ad = CD = ab
- 11. The side length of square ABCD is 12, e is a point on BC and be = 5, the vertical bisector Mn of AE intersects CD at point m and ab at point n, and the length of Mn is calculated
- 12. The area of a parallelogram is 86 square meters, which is the same as the area of a triangle with the same base and height______ ．
- 13. The top of a trapezoid is three times as big as the bottom. If the bottom is extended by 6 cm, it will become a parallelogram. How many cm are the top and bottom of the trapezoid?
- 14. The height of a trapezoid is 10 cm. If the upper bottom is increased by 6 cm and the lower bottom is increased by 4 cm, it becomes a parallelogram. Calculate the area of the original trapezoid
- 15. As shown in the figure, a, B, C and D are the four vertices of rectangle ABCD, ab = 25cm and ad = 8cm. The moving points P and Q start from point a and C at the same time. Point P moves to point B at the speed of 3cm / s until point B, and point Q moves to point d at the speed of 2cm / S. (1) P and Q, from the start to the second, PQ ‖ ad? (2) Question: P, Q two points, from the start to a few seconds, quadrilateral pbcq area of 84 square centimeters?
- 16. As shown in the figure, in the known trapezoidal ABCD, ad ‖ BC, ∠ B = 30 °, C = 60 °, ad = 4, ab = 33, then the length of the bottom BC is______ ．
- 17. In the right angle trapezoid ABCD, AD / / BC, ∠ B = 90 °, ad = 24cm, BC = 26cm, the moving point P moves rapidly from a along the side of ad to D at the speed of 1cm per second In the right angle trapezoid ABCD, AD / / BC, ∠ B = 90 °, ad = 24cm, BC = 26cm, the moving point P starts from a and moves at the speed of 1cm / s along the ad side to D, the moving point Q starts from point C and moves at the speed of 3cm / s along the CB side to B, and P and Q start from points a and C at the same time. When one point reaches the end point, the other point also stops moving?
- 18. As shown in the figure, in the diamond ABCD, ab = 2, ∠ C = 60 °, the diamond ABCD rolls to the right on the straight line l without sliding. Each rotation of 60 ° around a vertex is called an operation. After 36 such operations, the total length of the path that the diamond center o passes through is (the result retains π)______ ．
- 19. The ratio of the top to the bottom of a trapezoid is 3:5. The top is 1 meter shorter than the bottom, and the bottom is 25% higher. The area of the trapezoid is ()
- 20. As shown in the figure, in rectangular ABCD, e is a point on CD, angle DEA = 30 degrees, and AE = AB = 8cm, find 1, degree of angle EBC, 2, area of triangle Abe