In △ MNB, BN = 6, points a, C, D are on MB, Nb, Mn respectively, and the quadrilateral ABCD is a parallelogram, and ∠ NDC = ∠ MDA, then the circumference of quadrilateral ABCD is () A. 24 B. 18 C. 16 D. 12

In △ MNB, BN = 6, points a, C, D are on MB, Nb, Mn respectively, and the quadrilateral ABCD is a parallelogram, and ∠ NDC = ∠ MDA, then the circumference of quadrilateral ABCD is () A. 24 B. 18 C. 16 D. 12

In the parallelogram ABCD, CD ∥ AB, ad ∥ BC,
∴∠M=∠NDC，∠N=∠MDA，
∵∠NDC=∠MDA，
∴∠M=∠N=∠NDC=∠MDA，
∴MB=BN=6，CD=CN，AD=MA，
The circumference of quadrilateral ABCD = AB + BC + CD + ad = ma + AB + BC + CN = MB + BN = 2bn = 12
Therefore, D

As shown in the figure, m and N are the midpoint of AD and BC respectively, and P and Q are the midpoint of BM and DN respectively (1) Verification: △ MBA ≌ △ NDC; (2) What kind of special quadrilateral is quadrilateral mpnq? Please state the reasons

It is proved that (1) the rectangle is a quadrilateral,
∴AB=CD，AD=BC，∠A=∠C=90°，
∵ in rectangular ABCD, m and N are the midpoint of AD and BC respectively,
∴AM=1
2AD，CN=1
2BC，
∴AM=CN，
In △ mAb and △ NDC,
A kind of
AB＝CD
∠A＝∠C＝90°
AM＝CN ，
∴△MBA≌△NDC（SAS）；
(2) The quadrilateral mpnq is a diamond
The reasons are as follows: connect AP, Mn,
Then the quadrilateral abnm is a rectangle,
∵ an and BM are equally divided,
Then a, P, n are on the same line,
Easy evidence: △ ABN ≌ △ BAM,
∴AN=BM，
∵△MAB≌△NDC，
∴BM=DN，
∵ P and Q are the midpoint of BM and DN respectively,
∴PM=NQ，
A kind of
DM＝BN
DQ＝BP
∠MDQ＝∠NBP ，
∴△MQD≌△NPB（SAS）．
The quadrilateral mpnq is a parallelogram,
∵ m is the mid point of AD, q is the midpoint of DN,
∴MQ=1
2AN，
∴MQ=1
2BM，
∵MP=1
2BM，
∴MP=MQ，
The parallelogram mqnp is a diamond

As shown in the figure, m and N are the midpoint of AD and BC respectively, and P and Q are the midpoint of BM and DN respectively (1) Verification: △ MBA ≌ △ NDC; (2) What kind of special quadrilateral is quadrilateral mpnq? Please state the reasons

It is proved that (1) the rectangle is a quadrilateral,
∴AB=CD，AD=BC，∠A=∠C=90°，
∵ in rectangular ABCD, m and N are the midpoint of AD and BC respectively,
∴AM=1
2AD，CN=1
2BC，
∴AM=CN，
In △ mAb and △ NDC,
A kind of
AB＝CD
∠A＝∠C＝90°
AM＝CN ，
∴△MBA≌△NDC（SAS）；
(2) The quadrilateral mpnq is a diamond
The reasons are as follows: connect AP, Mn,
Then the quadrilateral abnm is a rectangle,
∵ an and BM are equally divided,
Then a, P, n are on the same line,
Easy evidence: △ ABN ≌ △ BAM,
∴AN=BM，
∵△MAB≌△NDC，
∴BM=DN，
∵ P and Q are the midpoint of BM and DN respectively,
∴PM=NQ，
A kind of
DM＝BN
DQ＝BP
∠MDQ＝∠NBP ，
∴△MQD≌△NPB（SAS）．
The quadrilateral mpnq is a parallelogram,
∵ m is the mid point of AD, q is the midpoint of DN,
∴MQ=1
2AN，
∴MQ=1
2BM，
∵MP=1
2BM，
∴MP=MQ，
The parallelogram mqnp is a diamond

As shown in the figure, ∠ ABC = 60 ° and ab = BC, ∠ man = 60 ° in ▱ ABCD, please explore the quantitative relationship between BM, DN and AB, and prove your conclusion

The quantitative relationship is BM + DN = ab,
Proof: link AC,
∵ ABC = 60 ° and ab = BC,
ν Δ ABC is an equilateral triangle,
∴∠BAC=60°，AC=AB，
∵ quadrilateral ABCD is a parallelogram,
∴AB∥CD，AB=CD，
∴∠ACD=∠BAC=60°，
∵∠MAN=60°，
∴∠BAM=∠CAN，
In △ ABM and △ can,
∠BAM＝∠CAN
AB＝AC
∠B＝∠ACN＝60° ，
∴△ABM≌△CAN（ASA），
∴BM=CN，
∴BM+DN=CD=AB．

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(1) It is proved that: the quadrilateral ABCD is a diamond,
∴∠BAC=∠DAC．
And ∵ EF ⊥ AC,
/ / AC is the vertical bisector of EM,
∴AE=AM，
∵AE=AM=1
2AB=1
2AD，
∴AM=DM．
（2）∵AB∥CD，
∴∠AEM=∠F．
And ? FMD = ∠ ame,  ame = ∠ AEM,
∴∠FMD=∠F，
The △ DFM is an isosceles triangle,
∴DF=DM=1
2AD．
∴AD=4．
The circumference of diamond ABCD is 16

It is known that, as shown in the figure, the quadrilateral ABCD is a rhombic. The vertical line EF of AC is drawn through the midpoint e of AB, intersecting ad at point m, the extension of intersection CD at point F, and the perpendicular foot is o Results: (1) m is the midpoint of AD; （2）DF=1 2CD．

It is proved that: (1) connect BD,
∵ the quadrilateral ABCD is a diamond,
Ψ Ao bisection ∠ bad, AC ⊥ BD,
∵ EF ⊥ AC, point E is the midpoint of ab,
The EM is the median line of △ abd,
ν m is the midpoint of AD;
(2) In △ ame and △ DMF,
∵∠EAM=∠FDM，AM=DM，∠AME=∠DMF，
∴△AME≌△DMF，
∴DF=AE，
∵AE=1
2AB＝1
2CD，
∴DF=1
2CD．

It is known that e is the midpoint of ad on the edge of diamond ABCD, and EF intersects AB at M perpendicular to AC, indicating that M is ab and is the midpoint

∵ the quadrilateral ABCD is a diamond
AC is diagonal
∴∠BAC=∠CAD
∵EF⊥AC
∴∠AFM=∠AEF
In △ AFM and △ AFE
∠BAC=∠CAD
AF=AF
∠AFM=∠AEF
The △ AFM congruent △ AFE
∴AM=AE
∵AB=AD
Point E is the midpoint of AD
The point m is the midpoint of ab

As shown in the figure, take any point m on the side BC of the square ABCD, pass through the point C as cn ⊥ DM, and cross AB to N. let the intersection point of square diagonal be o, try to determine the relationship between OM and on, and explain the reasons

 DCCM, ? is  DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM, ? DCCM,  DCCM

As shown in the figure, in trapezoid ABCD, ad ‖ BC, ∠ B = 90 °, ad = a, BC = B, DC = a + B, and b > A, point m is the midpoint of AB edge (1) Confirmation: cm ⊥ DM; (2) Find the distance from the point m to the edge of CD

It is proved that: (1) extending DM, CB intersecting at point E. (as shown in Fig. 1) ∵ in trapezoid ABCD, ad ∵ BC,  ADM =  BEM, ∵ point m is the midpoint of AB edge,  am = BM. In △ ADM and △ BEM,

In trapezoidal ABCD, ad ∥ BC, M is the midpoint of AB, AD + BC = CD, indicating DM ⊥ cm

Do me ∥ ad, hand over DC to E
Since m is the midpoint, me is the trapezoidal median line
So 2me = AD + BC
Because AD + BC = CD, 2me = CD
Because of the median line, de = EC = 1 / 2dc
In this way, de = EC = me can be obtained
So ∠ DME = ∠ mde ∠ EMC = ∠ MCE
The four corners above add up to 180, so ∠ DME + ∠ CME = 90
So vertical