8th grade math, clear steps!

Proof : Extend Fe to point M (prove that angular MEA = angular EFG) because FG = CG

So the angle GFC = angle c

And because ab=dc

So angle b = angle c

So angular b = angular gfc

Angle B + Angle BEF + Angle BFE = 180

Angular GFC + angular GFE + angular BFE = 180

So the angle efg = angle bef = angle mea

So AB is parallel to GF

And because ae=gf

So the quadrilateral AEFG is a parallelogram.

2) Angular fgc+2 angular GFC=180

and 2 angles FBE+2 angles GFC=180

Angular fbe + angular GFC = 90

So the angle efg = 90

And because AEFG is a parallelogram.

So the AEFG is rectangular.

Prove: (1) Because ab=cd, the angle b=anglec, because gf=gc, so the anglegfc=anglec=angleb, so ab gf, ae=gf, so the quadrilateral, aegg, is a parallelogram.

2) Go over G to do GM BC. Because the quadrilateral aefg is a parallelogram, b= gfhBecause gf=gc,gm fc so fgm= cgm=1 2( fgc)= efb.

So fgm+ gfm= efb+ gfm=90° and the quadrilateral aefg is a parallelogram. So the quadrilateral AEFG is rectangular.

If the water velocity is v, then the speed of the ship is 20v, x (19v) + x (21v)] 2x 20v) = 400 399, and the time taken by the boat to go back and forth between A and B in the river is 400 399 times the time taken by the boat to go back and forth between A and B in still water.

Solution: If the water velocity is V, the ship speed is 20V;

The time taken to go back and forth between A and B in still water is t1, and the time taken to go back and forth between A and B in flowing water is t2;

Side find: t2 t1;

t2=x/(19v)+x/(21v);

t1=2x/20;

obtained: t2 t1=400 399;

Answer: The time taken by a boat to go back and forth between A and B in the river is 400 399 times the time taken by a boat to go back and forth between A and B in still water.

The figure holds the sedan pants section and the fan is noisy.

<> please ask the wax liquid to pick the wheel and tell the ants.

The source of the solution is like Wang Lao in the state of Jianling:

Refer to the following pure early examinations: Brief Writings.

d=90°,ef cd,eg ad, efdg is a rectangular connection de,, de=fg

ABCD is a square, BC=CD, BCA= DCA=45°, CE=CE

cbe≌δcde，be=de，∴be=fg

The perpendicular lines EH and EI are made to AB and BC respectively through E, because AE is a square diagonal, we can know that EH=EG, EI=EF=HB, and the triangle HEB=EGF, so BE=FG