Connect PE and expand, pay BC to G.

Then PD = 2/3pf and PE = 2/3pg.

So DE is parallel to FG, and DE=2/3FG.

Because D and E are the center of gravity of the triangle, and F and G are the midpoint of AB and BC.

So FG is connected in parallel with AC, so DE is connected in parallel with AC, FG= 1/2AC.

So DE= 1/3AC.

12, (1)ADF area =1/2 * ad * af * sindaf = 3/2 * 20.5.

So sindaf = (2 0.5)/2

So DAF=45 degrees

Because AF is parallel to BE.

So the angle between AD and BE is 45 degrees.

(2) As can BE seen from the figure, the distance between AD and BE is the length of AB.

Because ab * ad+ab * af = 25 cm 2.

So AB = 5 cm

So the distance between AD and be is 5cm.

13, (1) Take the AC midpoint G and connect EG and FG, then EG =1/2bc =1/2ad = FG.

Since EF = 2 0.5/2 * AD, EF = 2 0.5 * EG = 2 0.5 * FG.

So EGF = 90 degrees, so BC and AD made a 90-degree deal.

(2) In the same way, ef = 3 0.5 * eg = 3 0.5 * fg.

So EGF= 120 degrees, so BC and AD have a deal 120 degrees.

14, (1) connects MB, MD 1, and it is easy to know that MB=MD 1.

Because o is the midpoint of BD 1, MO is perpendicular to BD 1.

Connecting OA and OA 1, it is easy to know that OA=OB=OD 1=OA 1.

Because m is the midpoint of AA 1, MO is perpendicular to AA 1.

(2) MO from (1) is perpendicular to BD 1 and AA 1.

So the distance from AA 1 to BD 1 is the length of MO.

In the triangle BMD 1, BM = 5 0.5/2 * a, BD 1 = 3 0.5 * a, so Bo = 30.5/2 * a..

So om = 2 0.5/2 * a, so the distance from AA 1 to BD 1 is 2 0.5/2 * a.

15. Take point G at A 1B 1 so that b1g =1/4 * a1b1,then FG and A/kloc-0.

C 1g = 17 0.5/4 * AB, then the number of C 1FG can be obtained from the cosine theorem, that is, A 1E and C 1F form an angle.

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