I can t do math and geometry problems in the second year of junior high school!! Seek high speed!! G

Proof of: (1) It is known that ADC=so AD=90° so 2)RT ABC, AC=6, BC=ABC=AC24-36=108, so BC=6 3

If you can't square it, you can change it yourself.

Evidence: (1) Because the angle ADC = angle B + angle DAB

Angle ADC = 2 Angle B So: Angle DAB = Angle B

And because: ad bisects the angle bac so: angle cad = angle dab thus: angle cad = angle dab = angle b

Because: angle c = 90 degrees so: angle cad = angle dab = angle b = 30 degrees.

2) Because: angle c = 90 degrees angle b = 30 degrees ac = 6 so: ab = 12 (the right-angled side of 30 degrees is equal to half of the hypotenuse).

Therefore: BC2=AB2-AC2=144-36=108 BC=6 root number 3

Solution: Because AD bisects BAC, ADC=2 B, so CAD= DAB= ADC- B=2 B- B= B, and B+ CAB=90°, so B=90° 3=30°, because B=30°, AC=6, so AB=12, and BC=6 3

It's okay to be stupid, usually you do less questions, and you can know the answer at a glance if you do more, and you don't have to think about it at all.

1) When PQ BC, the triangle APQ is similar to the triangle ABC, so.

There is 2t :(5-t)=4:5, and the solution, t =10 72) is pd as pd perpendicular bc, if quadrilateral pqp'C is a rhombus, then PD bisects QC perpendicularly, so AD = 4-(4-2T) 2 = 2+T PD:

bc=ap：ab pd：3= （5-t）：

5, so. PD=3(5-T) 5 Because AD:AC=PD:BC, so (2+T):4 = 3(5-T) 5:3

The solution is that t= 10 9 so pd= 7 3 , qd= 2-t =8 9 , and the Pythagorean theorem can be used to find pq= root 505 9 (the result is not necessarily right, but the idea is right, you reason and verify it again).

Is it on the math resource?,If so, hit the resource and evaluation answer on Google and the first answer is。 Ultra-complete.

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.

Extend aq, bc, let aq, bc hand over e

Because the angle d = angle dce, dq = cq, and the angle aqd = angle cqe, the triangle adq is equal to the triangle ceq

So ad=ce

So ap=cp+cd=cp+ce=pe

So angular pae = angular pea

So angular dae = angular pea = angular pae

So AQ is the angular bisector of the angle DAP.

It can't be a diamond.

The reason is: it is easy to know that the certificate is complete. Well, there is no such thing as what you say, because< ecf = 90 degrees, it has always been, and it is easier to prove from the negation that "ef equals cf".

It's not just a picture that you can see.

Is this definitely going to work? You don't need to think about this kind of question to know that it is, if it is not, the person who wrote the question has a disease in the brain.

Of course, when point O coincides with AC, it is a straight line, and point O is trapezoidal no matter how it moves!

Wow oh n years ago's problem.,It's so familiar.,I think I liked math the most.。。。

Although now I forgot about ==...

However, children's shoes ... It's better to ask if you don't understand.。。。

(1) Because the parabola y=-1 2x 2+5 2x-2 intersects with the x-axis at points a,b

So 0=-1 2x 2+5 2x-2 solves the equation and gives , x1=1 x2=4

So a(1,0), b(4,0).

Because the abscissa of point c is 0, that is, x=0, bring in the parabola to find the coordinates of c (0,-2) as the auxiliary line af vertical cd af=2

And because cd=cf+fd d(5,0) trapezoidal abcd area is (3+5)x2 2=8, I wrote ,,, wrong, no matter what.