How can I learn the math proof problems in junior 3?

Practice more, think more. Pay attention to textbooks and better reference books.

How to solve the junior high school mathematics certificateIn the junior high school mathematics and geometry learning, how to add auxiliary lines is a headache for many students, and many students often have difficulty in solving problems due to improper addition of auxiliary lines. The following are some common guide line practices that have been compiled into some "slippery" songs.

Everyone says that geometry is difficult, and the difficulty lies in the auxiliary line. How do I add an auxiliary line? Grasp theorems and concepts.

It is also necessary to study assiduously and find out the rules based on experience. There are angular bisectors in the diagram, which can be perpendicular to both sides.

Angles bisector parallel lines, isosceles triangles to add. Angular bisector line plus perpendicular line, three lines in one to try.

The line segment bisects the line vertically, often connecting the lines to both ends. There are two midpoints in the triangle, and when they are connected, they form a median line.

There is a midline in the triangle, and the extension of the midline is an isomidline. A parallelogram appears, symmetrically centrically bisecting points.

Make a high line inside the trapezoid, and try to pan it around the waist. It is common to move diagonal lines in parallel and make up triangles.

The certificate is similar, than the line segment, and it is customary to add parallel lines. For equal area sub-proportional exchange, it is very important to find line segments.

It is directly proved that there is difficulty, and the same amount of substitution is less troublesome. A high line is made above the hypotenuse, and a large piece of the middle item is proportional.

The radius is calculated with the chord length, and the chord centroid distance comes to the intermediate station. If there are all lines on the circle, the tangent points are connected with the radius of the center of the circle.

The Pythagorean theorem is the most convenient for the calculation of the tangent length. To prove that it is a tangent, the radius perpendicular line is carefully identified.

It is a diameter and forms a semicircle and wants to form a right-angle diameter chord. The arc has a midpoint and a central circle, and the vertical diameter theorem should be memorized.

The two chords on the periphery of the corner, the diameter and the end of the chord are connected. The string is cut to the edge of the tangent string, and the same arc is diagonally to the end.

If you encounter intersecting circles, don't forget to make common chords. Two circles tangent inside and outside, passing through the tangent point of the tangent line.

If you add a connecting line, the tangent point must be on it. The auxiliary line is a dotted line, and you should be careful not to change it when drawing.

Basic drawing is very important, and you must be proficient in mastering it at all times. It is necessary to be more attentive to solving problems, and often summarize the methods.

Don't blindly add lines, and the method should be flexible and changeable. Analyze and choose comprehensive methods, no matter how many difficulties there are, they will be reduced.

With an open mind and hard work, the grades rose into a straight line. Clear, what are the skills.

Proof: d, f is the midpoint of ab, ca, so df bc, because e is the midpoint of bc, so ec = bc, so df ec, so, the quadrilateral decf is a parallelogram because ac = 10, bc = 14So the quadrilateral decf = 2 (cf + ec) = 2 (5 + 7) = 24

The circumference is 24... 10+14=24

According to the median line, a parallelogram can be demonstrated. Get be=df

de=af equivalent substitution.

The perimeter can be calculated de+ec+fc+df=be+ec+cf+af=10+14=24

AC to O

In ADE and CBF.

ad=cb (parallelograms are equal on opposite sides).

ade=∠cbf

de=bf△ade≌△cbf(

ae=cfdae=∠bcf

and dac= BCA (two lines are parallel, with equal angles) ade cbf

ae=cf∠eac=∠dac-∠dae

fca=∠bca-∠bcf

dae=∠bcf

fca=∠dae

In EOF and FCO.

fca=∠dae

EOA = COF (equal to the vertex angle).

ea=cf△eof≌afco(

eo=ofao=oc

The rhombus has a circumference of 20

So side length = 5

The ratio of the two adjacent angles is 1:2, and they complement each other, so the angles are 60° and 120°

Draw 2 diagonal lines to get 2 equilateral triangles, side length = 5, area = 25 3 4 area of the diamond = 25 3 2

The point d, de ab, df ac on the isosceles triangle abc, ab = ac, bc, df ac with ad, then, the area of abc = ab*de 2 + ac*df 2 = (de+df)*ab 2

And the area of abc = height on the waist * ab 2

So, de+df = height on the waist.

That is, the sum of the distances from any point to the two waists at the base edge of the isosceles triangle is constant: the height on the waist.

Make a perpendicular line at any point on the bottom edge to the two waists, and the angle between the perpendicular line and the bottom edge is equal (because the two bottom angles of the isosceles triangle are equal) set to a, then the sum of the distances to the two waists is l1*cosa+l2*cosa=(l1+l2)*cosa

l1+l2 is the length of the base edge, which is constant, and a=180-b (b is the base angle) is also constant.

Therefore, the sum of the distances from any point to the two waists at the base edge of the isosceles triangle is a constant independent of the position of the point.

Another:

First question:

ABCD is square, ACB 45°, DCE 90°.

CF bisects DCE, DCF 45°.

Acf 180° ABC ECF 180° 45° 45° 90°. Combined with PA PF, we get:

a, p, c, f are concentric, afc acb 45°, paf is an isosceles right triangle with af as the hypotenuse, ap pf.

Second question:

When ap ag, combined with ab ad, we get: rt abp rt adg, bp dg, and then cp cg.

At this time, CPG is an isosceles right triangle with PG as the hypotenuse, CPG 45° ECF, PG CF.

i.e.: when ap ag, pg cf.

Area of PG CF, CPG Area of PGF.

Area of APG Area of PAF Area of PGF Area of PAF Area of CPG.

AB 2, PB 1, AP 2 AB 2 PB 2 4 1 5, Area of PAF AP 2 2 5 2.

Obviously there are: pc cg 1, cpg area pc 2 2 1 2.

Area of APG Area of PAF Area of CPG 5 2 1 2 2.

That is, when ap ag, the area of the apg is 2.

1) Connect AC, AC is ABCD diagonal, AC CFP, C is on a circle with AF as the diameter, FCE=PAF=45° PAF is an isosceles right triangle, PA=PF

2)∵pa=pg

abp≌△adg==>pb=gd==>gc=cp==>ac⊥gppg//cf

p is a point on bc, the position of p is uncertain, and s(apg) is also uncertain.

Let pb=x, then pc=2-x

s(⊿apg)=s(abcd)-2s(⊿apb)-s(⊿pgc)=4-2x-1/2(2-x)^2=2-1/2x^2

Because AB CD, and CF is an extension of CD, AB CF, and the quadrilateral ABC is trapezoidal. Furthermore, because E is the midpoint of BC and the point E is the intersection of the diagonal of the trapezoidal ABFC, the quadrilateral ABFC is a parallelogram.

1) aeb= fec, ab cd, abe= fcebe=ec triangle abe is all equal to triangle fce

ab=cf2) ab cd, and ab=cf quadrilateral abfc is a parallelogram.

1) Proof : ab parallel cd, then: abe= fce; ∠bae=∠cfe.

and be=ce, then abe δfce(aas), get: ab=cf

2) The quadrilateral ABFC is a parallelogram.

Proof: ab=cf, ab parallel cf, so the quadrilateral abfc is a parallelogram.

A set of quadrilaterals with opposite sides parallel and equal is a parallelogram).

1.Prove abe fce, then get ab = cf parallelogram, from ab cf, get baf = cfa (equal internal wrong angles), ac bf, quadrilateral acbf is a parallelogram.

2.parallel, or because the internal misalignment angles are equal. ( bef= dfe) Remember, you have to bring a picture in the future, otherwise it will be difficult to understand.

Certificate: In parallelogram ABCD.

Angular AOE = Angular COF

Angular OAE = Angular OCF

ao=co (diagonal bisect).

So the triangle aoe is all equal to the triangle cof

oe=of

2x²+(m+8)x+m+5=0

Discriminant formula =(m+8) -8(m+5)=m +8m+24=(m+4) +8>0

0, i.e., there are two unequal real roots.

Whether it's right or not, you can look at my train of thought first, no matter what the value of m is, it means that this equation has nothing to do with m, so if you put the equation about m together, and move the term to get m(x+1)+2x 2+8x+5=0, which has nothing to do with m, then m=0, 2x 2+8x+5=0, from the discriminant formula of the root=8 2-4*2*5=24>0, so there are two different real roots.