Junior 2 math problems triangles, junior 2 math problems, about triangles

Proof is that the connection CE, AD bisects the angle BAC and DC perpendicular AC, DE is perpendicular to AB Angle CAD=angle EAD, angle ADC= angle AD=AD The triangle ACD is all equal to the triangle AED AC=AEconnects the CE angle AD at point F AC=AE, the angle CAF = the angle EAF, AF=AF The triangle ACF is fully equal to the triangle AEF Angle AFC=Angle AFD=90°; CF=EF AD is the perpendicular bisector of CE.

ad=ae, so the angle ade is equal to the angle aedAnd because the angle DAF is equal to the angle EAG, the triangle ADF is all equal to the triangle AEGAF is equal to AG, so angle AFG is equal to angle AGF, and because angle ABC is equal to angle ACB, so angle AFG is equal to angle ABC, (the sum of the first two angles and the last two angles is equal and each pair of angles is equal).

So fg parallel bc, so de parallel bc

Similarity From the meaning of the title, it can be seen that acf= gca square abcd=>ac=root number2*cd square cdef=>cd=cf So: ac=root number2*cf also know cf=fg (from the question) so cg=2*cf=root number2*ac then:

ac cf=cg ac=root number2 at the same time: acf= gca so: triangle acf is similar to triangle gca then:

cag= 2 also know: 1 cag= acb=45° so: 1 2=45°

Let the three heights be H1, H2, and H3 respectively

Since half of the product of the base and height is the area of this triangle.

Therefore the product must be the same.

ah1=bh2=ch3

a:b:c=2:3:4, so a=2k, b=3k, c=4k2kh1=3kH2=4kH3

h1:h2=3:2=6:4

h1:h3=2:1=6:3

Therefore h1:h2:h3=6:4:3 looks.

Make a line parallel to the BC line to the right at point A, and draw a line parallel to the AB line, and the focus of the two lines is point D.

Make a parallel line to the left of point A, and draw a line parallel to the line AC, and the focus of the two lines is point D.

Make a parallel line to the right of point C, and draw a line parallel to the line of AC, and the focus of the two lines is point D.

It is the theorem that parallelograms are parallel to sides.

There are three dots, and these three dots are diagonal a, b, and c!

There are three points in total.

Here's how.

1.With ab and bc as the adjacent sides of the parallelogram, then the other point is to the right of ac 2 with ac and bc as the adjacent edge, then the other point is on the left side of ab 3AB and AC are adjacent to each other, and the other point is below BC.

Passing a is a parallel line of BC, and there are two points on this line.

Passing C as a parallel line of AB, there are two points on this line.

Passing through b is a parallel line of ac, and there are two points on this line.

And then there are three that are repeated, so there are three.

"'If there are three points, then these three angles should be diagonal to ab c'! ”

Because: af is so: because so: because: so: so: ce=cf

So: CEF is an isosceles triangle.

The intersection of AB and ED is O

Because abc is an equilateral triangle and d is the midpoint, dab=30° (three lines in one).

So bae=30°

Prove AEO ADO with SAS

oe=odaoe=

Make an extension line along BC to O, so that the triangle AOB is a right triangle, so that the length of all sides can be obtained through the triangle with special angles, because the angle ACO is 45 degrees, so AO=CO=root number 2, and the angle ABO is 30 degrees, so AB=2AO=2 root number 2

Bo is also equal to root number 6, so bc = root number 6 - root number 2

Analysis: 1Equality relational deformation.

2.Multiply 2 recipes on both sides.

3.Rotate transforms.

4.Question 3 will definitely be question 4.

5.(1) Isosceles triangle with three lines in one.

2) Do the BC perpendicular line over A.

3) Question (2) will definitely be question (3).

Do it yourself!!

If ABC is a regular triangle with a side length of A, connecting PA, PB, and PC, the ABC area is S, the PAB area is S1, the PBC area is S2, and the PCA area is S3, then S1+S2+S3=S,1 2A·PM+1 2A·pH+1 2A·PN=1 2A· (3A2), PM+PH+PN=(32)A(fixed value).

If abc is not equilateral, then pm+pc+pn is not a fixed value!