To prove geometry, it is best to use geometric methods to solve them, rather than analytic geometry.

It's a weird question.,And it's not a printed type.,I always feel that it's not ** not very accurate.,Look at the conditions and the goal of proof.,You just need to prove that AB is equal to df.,But that 45 degrees don't know how to use it.,Can't get into a right triangle.。

AD is the bisector of BAC, 1 = 2, in ADG and ADF, agaf

adadag=af 1= 2ad=ad, agd afd (Keiwa SAS)

agd=∠afd，dg=df

aed+ edf+ dfa+ fae=360°, edf+ bac=180°

AED+ AFD=180°, and 4+ AGD=180°, which is about 4= 3, de=dg, de=df

<> guide line diagrams. Proof: Dm ab is in m and dn ac is in n

Take the dme = dnf, dme + dnf = 180°

eaf+∠mdn=180°.

edf + eaf = 180°, mdn = edf (the complementary angle of the same angle is equal).

mde=∠ndf.

AD bisects BAC, dm=dn

Qi Yin MDE NDF(ASA).

de=df.

<> the point d is dm ab in m, dn ac in n, dme = dnf 90°

and ad ad bisect bac, and ad ad(common edge), amd and(aas), dm dn

EDF+ EAF 180, MED+ AFD 360-180 180, and Li Chi NFD+ AFD 180, MED NFD

In EMD and FND, Syntagion obtains: EMD FND(AAS), Hall De Disturbance State DF

There is indeed a lack of conditional conditions, and only the angular EDC can be calculated as 25 degrees.

Using the parallel line theorem, the wrong angles in the two parallel lines are equal and the adjacent inner angles are complementary.

(1) Positive thinking. For general simple topics, we can easily do them by thinking positively, so I won't go into detail here.

2) Reverse thinking. As the name suggests, it is to think about the problem from the opposite direction. The use of reverse thinking to solve problems can enable students to think about problems from different angles and directions, explore problem-solving methods, and broaden students' problem-solving ideas.

This method is recommended for students to master. In junior high school mathematics, reverse thinking is a very important way of thinking, which is more obvious in the proof questions, there are few knowledge points in mathematics, the key is how to use, for junior high school geometry proof questions, the best way to use is to use reverse thinking. If you are already in the third year of junior high school, you are not good at geometry, and you have no ideas for doing problems, then you must pay attention:

From now on, summarize the methods of doing the questions. After carefully reading the stem of a question, students don't know where to start, so it is recommended that you start with the conclusion. For example:

There can be a thought process like this: to prove that some two sides are equal, then combined with the graph, it can be seen that only two triangles need to be proved to be equal; To prove the congruence of the triangle, combined with the conditions given, see what conditions are still missing that need to be proved, and how to make auxiliary lines to prove this condition, and think about it like this, ......In this way, we can find the idea of solving the problem, and then write the process out correctly. This is a very useful method, so students should definitely give it a try.

3) Positive and negative combinations. In junior high school mathematics, the known conditions are generally used in the process of solving problems, so we can find ideas from the known conditions, such as giving us the midpoint of a certain side of the triangle, we have to think about whether to connect the median line, or whether to use the midpoint of the midpoint of the midpoint. To give us a trapezoid, we have to think about whether to make a height, or to translate the waist, or to translate the diagonal, or to complement the shape, and so on.

The combination of positive and negative is invincible.

Middle School Mathematics and Geometry Proof Skills, Proficiency in the use and memorization of the following principles is the key.

The first circle passes through the point o(0,0,0),a1(a1,0,0),c1(0,0,c1), so the center of the circle o1 is the midpoint of a1c1 (a1 2,0,c1 2), so the equation for a straight line that passes o1 and is perpendicular to the plane oxz is (x-a1 2) 0=y=(z-c1 2) 0①

Similarly, the equation for a straight line passing O2 perpendicular to the plane oxy is (x-a2 2) 0=(y-b2) 0=z②

Unless a1=a2 and the line has no common point, the proposition is not true.

Connect to CQ PQ

Then bpq is also an equilateral triangle bpq=60° in pqc, pq=pb= 3 qc=2 pc=1, so qpc=90°

So BPC= BPQ+ QPC=150°

Have you ever studied the cosine theorem?