It is known that in trapezoidal ABCD, ad ‖ BC, m and N are the midpoint of BD and AC respectively (as shown in the figure) Verification: (1) Mn ‖ BC; （2）MN=1 2（BC-AD）．

It is known that in trapezoidal ABCD, ad ‖ BC, m and N are the midpoint of BD and AC respectively (as shown in the figure) Verification: (1) Mn ‖ BC; （2）MN=1 2（BC-AD）．

(1) Proof: connect am and extend it, intersect BC at point E (as shown in Figure 2), ≓ ad ‖ BC, ≔ dam = ∠ BEM, ∠ ADM = ∠ EBM,

As shown in the figure: in trapezoidal ABCD, ad ‖ BC, ab = DC, m and N are the midpoint of AD and BC respectively, ad = 3, BC = 9, ∠ B = 45 °, then Mn = ___

Make me ‖ AB and MF ‖ CD and hand over BC to points E and f respectively,
Parallelogram Abem and parallelogram CDMF can be obtained,
∴BE=AM，FC=MD，∠MEN=∠B=45°，∠MFN=∠C=45°，
∴∠EMF=90°，
∵ m and N are the midpoint of AD and BC respectively,
∴AM=MD，BN=NC，
∴EN=NF，
‡ n is the midpoint of EF,
∴MN=1
2EF=1
2（BC-AD）=3，
So the answer is 3

It is known that in trapezoidal ABCD, ad ‖ BC, m and N are the midpoint of BD and AC respectively (as shown in the figure) Verification: (1) Mn ‖ BC; （2）MN=1 2（BC-AD）．

(1) Proof: connect am and extend it, intersect BC at point E (as shown in Figure 2), ≓ ad ‖ BC, ≔ dam = ∠ BEM, ∠ ADM = ∠ EBM,

It is known as shown in the figure: in trapezoidal ABCD, ab ‖ DC, points E and F are the midpoint of two waist AD and BC respectively   Proof: (1) EF ‖ ab ‖ DC; （2）EF=1 2（AB+DC）．

Connect AF and extend intersection BC at point G
∵AD∥BC
∴∠DAF=∠G，
In △ ADF and △ GCF,
∠DAF＝∠G
∠DFA＝∠CFG
DF＝FC
∴△ADF≌△GCF，
∴AF=FG，AD=CG．
∵ AE = EB,
∴EF∥BG，EF=1
2BG，
That is, EF ‖ ad ‖ BC, EF = 1
2（AD+BC）．

Quadrilateral ABCD, m and N are the midpoint of AD and BC respectively, GH intersects AB at g, intersects DC at h, GH ⊥ Mn, ab = DC, verification ∠ AGH = ∠ DHG

Set the extension line of nm to the extension line of Ba to P and the extension line of CD to Q,
Set the midpoint of BD as O and connect Mo and No
Mo parallel and equal to AB / 2, no parallel and equal to CD / 2,
AB=CD,MO=NO,∠OMN=∠ONM,
∠AGH+∠OMN=∠AGH+∠GPN=90°,
∠DHG+∠ONM=∠DHG+∠HQN=90°,
∠AGH=∠DHG.
Can you complete the details yourself?

△ in ABC, a = π / 6, (1 + radical 3) C = 2B, find angle c, if vector CB is multiplied by vector CA = 1 + radical 3, find edge a.b.c

(1) By sine theorem
(1+√3)/2=b/c=sinB/sinC=sin(120°-C)/sinC=(√3/2*cosC+1/2*sinC)/sinC
∴tanC=1
∴C=45°
(2)
∵CB/CA=sinA/sinB=sin60°/sin75°=2√3/(√6+√2)
∵ vector CB * vector CA = CB * ca * COSC = 1 + √ 3
∴CB*CA=√6+√2
∴CB^2=2√3
That is, CB = 4 times √ 12 = a
In addition, B = 4 times √ 3 + 1 / (4 times √ 3) and C = 2 / (4 times √ 3) are calculated accordingly