Answers and processes, speed solutions, answers and processes, speed

Solution: If the perpendicular line of the bottom surface of the hillside crosses the bottom surface of the hillside at the point H, then FH is perpendicular to the bottom surface of the hillside, and then perpendicular to the bottom line; After passing the point H, the perpendicular line of the bottom line of the slope crosses the bottom line at the point G, then the bottom line is perpendicular to the surface FHG, and then perpendicular to FG, so the angle FGH is the inclination angle of the slope of the hill 30 degrees. The line segment fh is the value sought.

In the right triangle feg, fg=ef*sin60=100*(quadratic root number 3) 2=50* (quadratic root number 3).

In the right triangle FGH, FH = FG * sin30 = 50 * (quadratic root number 3) * (1 2) = 25 * (quadratic root number 3).

100 times the root number three by half equals fifty times the root number three.

FG is perpendicular to EG

fg = half of the root number 3 times 100 = 50 times the root number 3 in the right triangle fgh.

fh = 1 2fg = 25 times root number 3

(1) Prove: Because in the trapezoidal ABCD, AD BC, AB=DC, so the trapezoidal ABCD is an isosceles trapezoid, the angle B=Angle C, because GF=GC, so the angle GFC= angle C, so the angle B=Angle GF, so AE GF, and because AE=GF, the quadrilateral AEFG is a parallelogram.

2) Because GFC+ GCF=180° and GFC= EBF GFC= GCF so 2 EBF+ GFC=180° and because FGC=2 EFB so 2 EBF+2 EFB=180° with 2 gives EBF+ EFB=90° so AEB=90° so the quadrilateral AEFG is rectangular.

1.To prove that parallelograms can prove that the opposite sides are parallel and equal.

First of all, ae=gf, so as long as the certificates are parallel, it's fine.

Because it is an isosceles trapezoid, so the angle b = angle c and cg=gf so the triangle cgf is an isosceles triangle so the angle c = angle gfc

So the angle GFC = angle b so AB is parallel to GF i.e. AE is parallel to GF so AE is parallel and equal to GF.

That is, the quadrilateral AEFG is a parallelogram.

2.GFC+ GCF+ FGC=180° again. GFC= C and FGC=2 EFB So.

2 GFC+2 EFB = 180° then GFC+ EFB=90° so EGF=90°

then the quadrilateral AEFG is rectangular.

Proof: 1The bisector of AE BAC, the extension of EF AB at F, and the EG AC at point G

ef=eg and de are the perpendicular bisectors on the edge of bc, and the source is EB=EC EBF ECG(hl).

bf=cg2.Hail eggplant refers to EF=eg EAF EAF EAG(HL)af=AG 2AF=AF+AG=AB+BF+AC-CG=AB+ACAF=(AB+AC) NAHENG2