Help!! Mathematics and geometry problems in the second year of junior high school, urgent!! Mathemat

Because b 45 so hgb 45 so cge 45 because d 90 so gce 90 so ceg 45 so triangle cge is an isosceles right triangle i.e. ce cg, and because b 45 d 90 so agb is an isosceles right triangle i.e. ag=bg

Because ag=dc de=dc=ce,bc=bg=gc so bc=de

I'm just a few ideas, there are no complete steps, it's better to draw a picture to understand it more clearly.

Connect ag because of the perpendicular line so ag=bg has because the angle b=45 so the angle bag=45 so the angle bga is a right angle ag=dc=bg because the triangle gec is an isosceles triangle (45 according to the angle to the apex) so ce=cg middle and upper so dc+ce=bg+cg i.e. bc=de

1 If two figures are symmetrical with respect to a line, then the segments connecting the symmetry points are bisected perpendicularly.

3 There is a point at equal distances to the two endpoints of the line segment, and there is a vertical bisector of a line segment.

4 If one of the outer angles of an isosceles triangle is equal to 40°, then the degrees of the base angle of the isosceles triangle are

5 In the equilateral triangle ABC, AD is high on BC, then bad

6 The degree of the obtuse angle formed by the intersection of two high lines of an equilateral triangle is

7 If a string of numbers seen in the mirror is " ", then the string of numbers is

Yes is an isosceles triangle

2. Multiple choice questions: There are 8 sub-questions in this major question; 3 points for each question, a total of 24 points Among the four options given in each question, only one is correct, please fill in the letter before the correct answer in the parentheses after the question 3 points are awarded for each question if you choose correctly, and zero points are awarded for incorrect, non-selection or multiple choices

11 The following are the trademarks of the four major banks in China, among which the non-axisymmetric graphics are (

What's the use of making an extension cord?

Connecting AP, the EP of the easy triangle is equal to the triangle CFP, so PE=PF, so PEF is always an isosceles right triangle.

de bc, which is obtained by d= dcn, dem= n, dm=cm:

dem≌δcnm

em=mn,bm is the midline of RTΔEBN hypotenuse slag.

bm=mn, n=mbn.

Again cm=cb (posture ridge quietly by known condition cd=2cb), so mcn=2 mbc

Pass the M point as Mo Be, and connect to BM

be ad, mo be, so: mo parallel to ad and bc, dem ome, bmo cbm

Again: m is the midpoint of cd, so o is the midpoint of be, and two triangle bmos are equal to emo, so: beat and ome bmo

ab＝2ad

M is the midpoint of DC, so: BC cm, then the triangular base liquid staring at CBM is isosceles.

So CBM BMC

So:, dem ome bmo bmc and: ome bmo buried bmc emc so:: emc=3 dem

Extend the dust suspicion EM and BC intersect at point n

de bc, which is obtained by d= dcn, dem= n, dm=cm:

dem≌δcnm

em=mn, and BM is the midline of the rTδEBN hypotenuse.

bm=mn, n=mbn.

and cm=cb (by known condition cd=2cb), so mcn=2 mbc gets d= mcn=2 mbc=2 n=2 dememc= d+ dem=2 dem+ dem=3 dem. Sidley.

ABP is fully equal to ABQ

You know that, right? So bp=dq, because bc=dc, so pc=qc, so qpc makes an isosceles right triangle, let bp be x

cm, because pc=qc=5-x

So qp = 10-5 times the root number 3

You are a junior sophomore, you can use the Pythagorean theorem, I use the trigonometric source banging function) and then because qp=ap, so.

bp squared + ab squared = ap squared, so that hail can be suspected to get x, and then reed closed substitution into questioning.