Mathematics problems in the second year of junior high school related to the determination of paral

If the parallel line ed of BC intersects AC at point D, then CDEH is a parallelogram, EH=CD, because DE is parallel to BC and gets the angle AED=angle FBG, because GF is parallel to AC and gets the angle BFG=angle EAD, and because AE=BF so ADE and FGB are congruent, so FG=AD then AC=AD+DC=FG+EH

The D spot was added by myself. Let ad=fg).

Because ad=fg

a= gfb (because ac fg).

ae=bf, so ade fgb (sas), so aed= b

So de BC

So c= ade= deh

And because of EH FG

So che= cgf

Because ade= fgb

So 180°- ade=180°- fgb i.e. cde= cgf

Because c= deh

cde=∠cgf

Therefore, the quadrilateral cdeh is a parallelogram (because the diagonal is equal), so eh=cd

So ac=cd+ad=eh+fg

Because I came up with it myself, it may be a little troublesome, and I don't know what the standard answer is? Hope to give points - do not plagiarize.

Intercept ad equal to fg on ac, connect de, and all problems are solved.

It is easy to prove with SAS that the triangle ADE is all equal to the triangle FGB.

Congruent condition gets the angle ade = angle fgb = angle c

DE is obtained as parallel to BC (isotope angle) HE is parallel to AC (known), and DEHC is a parallelogram (two pairs of sides are parallel), HE=cd, AD=FG (known as a figure).

Get ac=eh+fg

Passing point E is made as an ED parallel BC to AC in D, by

They're all right, you can go for it!! Remember the conclusion: the line connecting the midpoints of the four sides of any quadrilateral is a parallelogram.

Connect the diagonals and use the median line theorem to get the conclusion you want to prove.

1 If one set of opposites of a quadrilateral is equal and the other set of opposites is parallel, then the quadrilateral is a parallelogram.

Is it? A: No, it is not. Because of the isosceles trapezoid.

It's a special quadrilateral.

2 If the two opposite sides of a quadrilateral are parallel to each other, is it a parallelogram? Be.

3 If the two sets of opposite sides of a quadrilateral dust shape are equal, then is it a parallel quadrilateral? Be.

1.Let the quadrilateral abqp be a parallelogram after x seconds.

18-3x=2x

x = perimeter is (

2.Let the quadrilateral pdcq be a parallelogram after x seconds.

10-2x=3x

x=2pd=qc=6cm

The circumference is (15+6)*2=42cm

E and F are the midpoints of Ca and CB = > EF AB = > EF AD (1).

D and F are the midpoints of BA and BC = > DF AC = > DF AE (2).

1) and (2) = > ADFE is a parallelogram = > DE and AF are bisected with each other.

Solution: Take the midpoint g of de and connect AG

Ad BC, AF BC at point F

adg=∠cbg ①

bae=∠afb-∠abf=∠afb-∠abc=90°-75°=15° ②

af ad and thus ag is the midline on the hypotenuse de of a right triangle.

ag=1/2de=gd=ge=ab

then dag= adg, agb= abg

Let adg=x

Then dag= gda=x, agb= abg= abc- cbg= abc- adg=75°-x [by adg= cbg].

AGB is the outer angle of the triangular ADG.

Thus agb= dag+ adg=x+x=2x

75°-x=2x is obtained by

x=25° thus abg= abc- cbg=75°-x=75°-25°=50°

From aed= bae+ abg=15°+50°=65°

1.The inner wrong angles are equal and the two straight lines are parallel, and because the two straight lines are equal, they are parallelograms.

2.The triangle abc and the triangle cda are congruent, so ad=bc, so the two sets of opposite sides are equal, so they are parallelograms.