How to solve the proof problem of junior high school mathematics I don t know where to start

In fact, the proof questions of the first and second years of junior high school are not very difficult to think about carefully! It's too late to go for a review!

Here's an example!

In the square ABCD, G is a point on the diagonal AC, connecting GB, GD, GE perpendicular to Cd to point E, GF perpendicular to Gb, intersecting Cd with point F, verifying: (1) ED=EF

2);cg=√2cf+ag1)

Over G is perpendicular to BC to P

It can be clearly seen that the quadrilateral gpce is a square so gp=ge, and the angle gpb = angle gef = 90 = angle pge

Because the angle pge=90, the angle pgf + the angle fge=90

Because GF is perpendicular to GB, the angle BGF=90, so the angle PGF+angle BGP=90

So the angle BGP = the angle fge

and gp=ge, angle gpb=anglegef=90

So triangle BGP is all equal to triangle fge

So BG=FG

Because ABCD is square and AC is diagonal.

So bc=dc, angular bcg=angular dcg, cg=cg

So the triangle bcg is all equal to the triangle dcg

So BG=GD

So gd=fd

and ge perpendicular to df to e

So ge is the midline of DF, de=fe

2) Pass F as Fq perpendicular to CD, and pass AC over Q

It is evident that CQ = 2CF

Then, just use the verification of ag=gq

In the isosceles right triangle CGE and the isosceles right triangle CAD, CG = 2CE and AC = 2CD can be seen

gq=cg-cq=√2ce-√2cf=√2（ce-cf）=√2ef

ag=ca-cg=√2dc-√2cfce=√2（dc-ce）=√2de

Because ef=de, gq=ag

So CG=CQ+GQ= 2CF+AG

And elementary school younger brothers (sisters), uh. Sister, I'm also in my first year of high school, come over! I know it's good that the bookstore has a book!

It's not a salesman) The author's name is Wang Houxiong! There are example questions in the book, analysis! And the title!

Whether it's the third year or the first year.

In the first and second years of junior high school, the analysis methods and solutions of the proof questions are similar. First of all, you should carefully read the known conditions of the question and the knowledge points involved in the associative conditions. If it is a geometric figure, draw the intention if necessary and mark the given conditions in the diagram, and then associate many conclusions that can be obtained according to the given conditions; Then take a look at the conclusion of this question, guess the knowledge points used to prove the conclusion, and finally analyze it in combination with the known conditions.

You should start with the simplest parallelogram and read the book slowly, and if you don't know how to ask the teacher, it's not a shameful thing.

First of all, you have to understand that the proof questions start from the parallelogram, and there are three ways to prove the parallelogram, which are 1, two groups of opposite sides are parallel 2, two groups of opposite sides are equal 3, and one group of opposite sides are parallel and equal, only if you understand the parallelogram, you can prove the diamond and rectangle, and on this basis, you can prove the square. (I won't say how to prove that it is a diamond, a rectangle, or a square, but you should have it in your textbook.) ）

When you do the question, for example, the question asks you to prove a square then you have to know if it is a rectangle or a diamond, if you don't know, just keep pushing forward to see if you know that it is a parallelogram, if you don't know yet, the question gives you enough conditions to prove that it is a parallelogram, and then slowly push back until you prove the square.

I also came from the third year of junior high school, and I just entered my first year of high school this year, I hope you will work hard, do not disappoint the expectations of your parents and teachers, and wish you to be admitted to your ideal high school.

You should first finish the content you have left behind, and then find a teacher or a good student to teach you simple proof questions, from which you slowly understand the solution method, if you want to learn it, you can only comprehend it with your heart, after all, learning is your own business, no matter how much others teach you, if you can't realize it, you still can't learn.

Just memorize the basic definitions, and then apply the known conditions of the proof questions.

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.

Solution: Because ABCD is diamond-shaped.

So ad=ab

And because E is the midpoint of AB.

So 2ae=ab=ad

And because de is perpendicular to ab

So the angle dae = 60 (in a right triangle the right angle of the 30° angle is half the right side of the hypotenuse) so the angle abc = 120

Because ab=4

So ae=2

So de=2 3

So s-abcd=8 3

ABCD is a parallelogram, AB=CD, AD=BC, angle A=angle CE, F is the midpoint, AE=CF

In ABEs and CDFs, AB=CD, CF=AE, and AC=ACabe CDF

df=be, and bf=de, bfde is a parallelogram, be df

I'm busy, so I'll tell you about my ideas.

With a triangle card.

The congruence of DCF and ABE is proven, so that DFC= AEB is obtained, and then the sum of the inner angles of the triangle ABEs is 180°, and because A+ ABF=180°, DFC= EBF is obtained

Then this uses the theorem that the isotope angles are equal and the two straight lines are parallel.

That's it.

2.Evidenced with parallelograms.

DE and BF are parallel and equal, so the quadrilateral DEBF is a parallelogram, and then it is natural to get be parallel to DF.

The parallelograms are parallel and equal to each other.

The midpoint is still parallel and equal.

A set of quadrilaterals that are parallel and equal to the opposite sides is a parallelogram.

So the opposite sides of the small parallelogram are also parallel.

over

(1) Proof of: b= b , bc=b c, bce= b cf, bce b cf;

2) Solution: ab is perpendicular to a b, for the following reasons:

The angle of rotation is equal to 30°, i.e. ecf = 30°, so fcb = 60°, and b = b = 60°, according to the inner angle of the quadrilateral and the degree of bob is known to be 360°-60°-60°-150°=90°, so ab is perpendicular to a b

Proof : b= b1, bc=b1c, cf is the common edge, bcf b1cf

2) a=30°, fcb1=90°-30°=60°, b1=60°, so cfb1= afo=60°, aof=90°

i.e. ab a1b1

∵∠acb=∠acb1,ac=ac,bac=∠b1a1c

Triangle congruence.

2.Because ace=30

So acb1=90-30=60

So cfb1= afa1=60

So aof=180-60-30=90

So ab is perpendicular to a1b1

I'll tell you, typing is a little slow, but first of all, the question is not super wrong?

What is the relationship between ABC and ABC in your 2P=A+B+C?

AD bisects bac, de ab, df, perpendicular ac, so de and df are the distances from the point on the bisector of the angle to both sides of this angle, respectively, so de=df

s△abc=(1/2)(ab*de+ac*df)=（1/2）(ab+ac)*de

So 28=(1 2)(20+8)*de

de=2cm

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:

Proof that d, e, and f are the midpoints of AB, BC, AC, respectively, and that de and df are the median lines of abc.

de ac, df bc, quadrilateral, decf are parallelograms.

d and f are the midpoints of the edges ab and ca, respectively.

DF is the median line of BC.

df is parallel and equal to 1 2bc

The quadrilateral decf is a parallelogram.

According to the median line theorem.

de∥cf df∥ce

So decf is a parallelogram.

DF and EC are parallel and equal, and DE and CF are parallel and equal.

My own summer vacation homework is still well written, pinched = =