How to get to the Railway Academy Additional 20 rewards

Starting Point: Qingyun Road.

Destination: Railway Institute.

Route 1: From Shanghai Hospital of Traditional Chinese Medicine, take 963 (Shanghai Hospital of Traditional Chinese Medicine - Chenhang), transfer to 744 (Shanghai Railway Station - Fengtang) at Shanghai Railway Station, and arrive at Railway University. Approx. km.

Route 2: From Zhijiang Middle Road, take 823 (Fengzhen Road - Shanghai Railway Station), transfer to 859 (Taopu Xincun - Jinyuan Road) at Gonghexin Road, Zhongxing Road, and arrive at Railway University. Approx. km.

Route 3: From Zhijiang Middle Road, take 823 (Fengzhen Road - Shanghai Railway Station), transfer to 517 (Shanghai Railway Station - Universal Park) at Shanghai Railway Station, and arrive at Railway University. Approx. km.

Route 4: From Zhijiang Middle Road, take 823 (Fengzhen Road - Shanghai Railway Station), transfer to 117 (Shanghai Railway Station - Taopu Wucun) at Shanghai Railway Station and arrive at Railway University. Approx. km.

Route 5: From Xibaoxing Road, take 829 (Weining Road - East Sports Club Road), transfer to Beijia Special Line (North District Bus Station - Jiading Passenger Transport Center) at Jiaotong Road, and arrive at Zhennan Road and Zhenbei Road. Approx. km.

Route 6: From Xibaoxing Road, take 829 (Weining Road - East Sports Club Road), transfer to Beijia Line (North District Bus Station - Jiading Passenger Transport Center) at Jiaotong Road, and arrive at Zhennan Road and Zhenbei Road. Approx. km.

Route 7: From Zhijiang Middle Road, take 823 (Fengzhen Road - Shanghai Railway Station), transfer to 107 (Liangcheng New Village - Wanli City) at Guangyue Road and arrive at Wanli City. Approx. km.

Route 8: From Xibaoxing Road, take 966 (Shiguang Xincun - Shanghai West Railway Station), transfer to 822 (Jinyuan Road - Universal Park) at Shanghai Railway Station, and arrive at Zhennan Road and Zhenbei Road. Approx. km.

Route 9: From Xibaoxing Road, take 966 (Shiguang New Village - Shanghai West Railway Station), transfer to 708 (Anyuan Road, Yanping Road - Taopu Qicun) at Wuning New Village, and arrive at Zhennan Road and Zhenbei Road. Approx. km.

Route 10: From Xibaoxing Road, take 966 (Shiguang Xincun - Shanghai West Railway Station), transfer to 838 (Taopu Xincun - Meili Garden) at Wuning New Village, and arrive at Tiedao University. Approx. km.

Where are you? Dalian?

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The auxiliary line of the landlord is very good, and the GM ef vertical bisect AB is set to the point M ef

ae＝be∠eab＝∠b＝30°

AEG EAB B 60° Bed Gem Gh GM (Equal distance from the point on the bisector of the angle to both sides of the angle) AG FG

rt△fgm≌rt△agh

efg＝∠eag

bgf＝∠bef－∠efg＝60°－∠efg＝60°－∠eag∠agc＝90°－∠cag＝90°－(cae－∠eag)＝90°－(30°－∠eag)＝60°＋∠eag

agf＝180°－(bgf＋∠agc)＝180°－(60°－∠eag＋60°＋∠eag)＝60°

ag fg afg is an equilateral triangle.

fg＝af＝bf

If you learn the four-point contour, it is even simpler, as long as you prove the EFG EAG, you can get the a, f, e, and g four-point contour.

agf＝∠aef＝60°

AFG is an equilateral triangle.

fg＝af＝bf

According to the inscription, bac=60°, ac=ab 2=ad, and acd is an equilateral triangle if cd is connected.

Remember m is the midpoint of af (mg is the perpendicular bisector of af), take the midpoint n of ad, pass m and n as a straight line, and know mn fd, fd ab, mn ab by the median line theorem of the triangle, and the straight line mn is the perpendicular bisector of ad.

ACD is an equilateral triangle, and the straight line Mn must pass through point C, then CN bisects ACD, ACM=30°.

AMG=90°, ACG=90°, AMGC is a circle with a quadrilateral, AGM= ACM=30°, mg is the perpendicular bisector of af, agf=30°+30°=60°, agf is an equilateral triangle, af=fg, fe is the perpendicular bisector of ab bf=af=fg.

First of all, D is the midpoint of AB, E is a point on BC, and De AB, F on the extension line of ED means that DF is the midpoint of AB, and there must be BF=AF

Then it is easy to know fg=ag from the af perpendicular line that passes g

In summary, if you want to prove bf=fg, you can prove af=ag

To get af=ag, just show that in an isosceles triangle afg there is an angle of 60 degrees.

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benefit to what is good, to followed by (object of benefit).

Benefit of benefit is a noun, e.g. benefit of learning English

1) Take a little E on AC so that CE=CP and PE because the angle ACB=60 degrees.

So the triangle PCE is an equilateral triangle.

cp=pe=ce

In the triangle APE vs. triangle PCD.

Angle AEP = 180 degrees - Angle CEP = 120 degrees.

Angle CPD = 120 degrees.

Angular AEP = Angular CPD

Angle ace = 60 degrees - angle ECD = angle DCP

So the triangle ape triangle pcd

ap=cd, then cp+cd=ae+ce=ac

2) The second question seems to be wrong.

If you look at the first diagram, the key to the first question is to find cd=bp

I am also a liberal arts student who has been recruited into the lake railway, and it is easy to fight quietly in the single-recruitment exam. The math test questions are basically the difficulty of the sample questions in the bridge banquet textbook, and Chinese and English should not be a problem for liberal arts students! The interview is not difficult, and it is good to behave normally without being nervous.

The single move mainly depends on the relationship, and if the empty slag has something to do with it, you can enter casually.

I'm from Shandong, and I don't know anything about this school. Ask someone else.

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Try registering the dll file, find the location of the file, and then enter the regsvr32 file location in cmd, e.g. regsvr32 d:.