Mathematics and Geometry in the third year of junior high school... Urgent.. speeding

Because the angle BOA = the angle OAD + the angle ODA

Because OA=OD, so angular OAD=angular ODA

That is, angular BOA = angular ODA + angular ODA

i.e. angular boa = angular ode

So OA parallel ed

Because AE is perpendicular to CD, the angle AEC = 90 degrees.

So the angle OAE = angle AEC = 90 degrees.

i.e. OA is perpendicular to AE

So AE is the tangent of the circle O.

Because ab cf

So the angle abf = the angle cfe

Because the eggplant mass is MN is the axis of symmetry.

So angular PBC = angular PCB

Because the angle abc = the angle dcb

So angular abf = angular dcp

So angular CFE = angular dcp

Because EFC is similar to PDC.

So there are two scenarios.

1: Angular ECF = Angular PDC

2: Angular cef = angular PDC

In the first case, pn = 2

The second scenario is explained below:

i.e. angular cef = angular pdc

Because the angle cef = the angle ped

So pde is an isosceles triangle, and the loose feast is pd=pe, and pn=xtan is set to pn=xtan angle, pbn=x

By the sine formula: BC sin angle bec = be sin angle ecbbe = punch Na silver (

bp=(x^2+

pd=((4-x)^2+

Substitution equation: pd=be-bp

Finally, we get: 8x 3-49x 2+87x-36=0x-3)(8x 2-25x+12)=0

Gets: x1=3, x2=, x3=

I've done this question before, and here's how to answer it.

In the triangle AMC and the triangle CNB, AC=BC, AMC= BNC=90 degrees, ACM+ NCB=90 degrees, NCB+ CBN=90 degrees, so ACM= CBN, RT AMC RT CNB, BN=cm, CM=BN, (1).

cn=am，（2）

1)-(2), cm-cn=bn-am, mn=cm-cn, mn=bn-am

Are you sure. This is the second question, alas. I guess it's an isosceles triangle. Otherwise, it can only be similar. And then it doesn't matter.

There is absolutely one condition missing: ac=bc

This is the topic of the second sub-question, and you go and look at the general topic.

oe=of

Proof : co ab, ce ad, afo= cfg, so c= a and because oa=oc, aof is all equal to coe(asa) so of=oe

oe=of.

Proof: OCE+ OEG=90;

oaf+∠oeg=90º.

oce = OAF (co-angles of the same angle are equal);

and oa=oc; ∠coe=∠aof=90º.

coe≌⊿aof(asa),oe=of.

oe=of

Prove that AOF is fully equal to COE

A pair of right angles, fao= eco, oa=oc, corner edge theorem.

Junior high school students. It's better to do it yourself, otherwise after you get to the high school entrance examination, you really look at your results and don't know the way out in the future.

lz, I need to practice more on this kind of topic. Asking for answers online won't help you.

oe=of

First find the angle cfg=afo, and then use the triangle congruence.

Dear, which one do you want to ask? 22 words ... What about the figure?

1. Intercept of=OP on OB, then ofp= OPF=45°, BFP=135°

PAQ= PAB+ BAQ=45+90=135° BFP= PAQ, BPQ=90°, FBA= APQ(Same as BPO), ob=OA, BF=PA (equal minus equal difference) BFP PAQ(ASA), BP=PQ2, as PD op in P, PD intersecting AQ in D, PD= pad=45°, PA=PD, BAP= QDP=135°, and BPA= QPD (same as BPD's co-angle), abp dqp(aas)

dq=ab,∵pm⊥aq,∴pm=dm

mq=dq+dm=ab+pm

i.e. mq-ab=pm

Experience: If you can't make up isosceles right triangles, you might as well try to make up the square, but it's the same if you don't make up.

First of all, it is noticed that the area of the rectangle is 2 times that of the triangle AED, so first find the area of the triangle AED de=4 3, l AED=90 degrees, L dae=60 degrees, so ae=de*cot(60)=DE3=4The area of the triangle AED: 1 2*de*ae=8 3 So, the area of the rectangle is: 16 3