Urgent!! A junior high school math problem!! P is a point on the bisector of AOB, PC OA, PD OB, and

The equal line segment is pc=pd oc=ob

p is a point on the aob bisector and pc oa, pd obpc=pb po cd is e

ce=de=1/2cd=4

There is the Pythagorean theorem to get pe=3

There is a proportional relationship between OC PC=CE PE

oc=20/3

pc=pd,oc=od,p is a point on the aob bisector and pc oa,pd ob

PC=PB PO CD is E

ce=de=1/2cd=4

There is the Pythagorean theorem to get pe=3

There is a proportional relationship between OC PC=CE PE

oc=20/3

pc=pd oa=ob

cd opopc area oc*pc=op*

oc^2+pc^2=op^2

oc=20/3

Or. Similar PCE (E is the intersection of CD Op) POC

1.Equal segments have od=oc, cp=dpBecause the distance from one point to both sides on the bisector of angles is equal by theorem, and because OPC and OPD have the same side op, and cop= dop, in these two right triangles, the two triangles are congruent, so oc=od

2.It can be seen that cd intersects op at point e, there is ce op, and ce=de=4, obtained by the pythagorean theorem, pe=3, then ce=oe ep, oe=16 3, so op=25 3, obtained by co cp=op ce, oc=(25 3 4) 5=20 3

1、pc=pd，oc=od。

2. Let the bisector line and CD intersect with the point E, and the Pythagorean theorem ep=3 has the ratio of similar triangles: 3 5=4 oc

oc=20/3

1、pc=pd，oc=od。

2. Let cd and op intersect at the point, then ce=1 2cd=4, and from the Pythagorean theorem, we can know that pe=3, then let oc=x, and the column equation:

25 + x squared) open root number = (x square - 16) open root number + 3, the solution is x = 20 3.

It can be obtained by using the AAS corner edge formula.

It depends on the situation, and the reason p is the moving point on the circumference that is different from the known sixth.

Forehead ......There is no picture, ah......I don't know where e is......

You can make up the downside, and it's OK to make up a square with Q as the center point, and you won't ask again.

1. Eggplant angle aoc=dof;

Angle AOC+COD=90, Angle DOF+COD=90;

angular aoc+dof;

2. Mutual co-angle.

aoc,cod）;(cod,dof);(dof,bof);(bof,aoc);

3. There are 3 pairs of complementary angles of the Zen only school;

They are (AOC, COB) He He; (aod,dob);(aof,fob)；

1) Proof of: paq= bac, paq- pac= bac- pac, i.e. bap= caq, in abp and acq, ab=ac bap= caqap=aq

abp≌△acq（sas），bp=cq；

From (1) abp acq, abp= acq, in abo and eco, aob= eoc, abp= acq, bac= bec, i.e. = ;

If the point p moves on the straight line ad (does not coincide with the point a), the quantitative relationship between and is equal or complementary, and the reason for the same is the same; The reason for complementarity is as follows: as shown in the figure, abp acq, abp= acq, acq= eco, abp= eco, eoc= aob, eco aob, ceo= oab, peq+ ceo=180°, peq+ bac=180°, i.e. + =180°

1) Conjecture: AP CQ proves as follows:

In ABP and CBQ, AB CB, BP BQ, ABC PBQ 60°, so BCQ can be seen as AP CQ obtained by rotating BAP 60° clockwise around point B

2) By PA PB PC 3 4 5, can set PA 3A, PB=4A, PC=5A

Connect PQ, in PBQ, since PBQ 4A, and PBQ 60° PBQ is a regular triangle PQ 4A

So in pqc, pq2 qc2 16a2+9a2=25a2=pc2

pqc is a right triangle, pqc=90°

1) Conjecture: ap=cq, prove: abp+ pbc=60°, qbc+ pbc=60°, abp= qbc

ab=bc,bp=bq, abp cbq,ap=cq;

2) From PA:PB:PC=3:4:5, you can set PA=3A, PB=4A, PC=5A, connect PQ, in PBQ.

Since pb=bq=4a, and pbq=60°, pbq is a regular triangle pq=4a

So in PQC.

PQ2+QC2=16A2+9A2=25A2=PC2 PQC is a right triangle

Because abc is equilateral, pbq=60

So abp=60-pbc=cbq, ab=bc because bp=bq

So ABP congruent CBQ SAS

So ap=cq

PQC is a right-angled triangle.

Because bp=bq, pa=cq

So cq:bq:pc=3:4:5

And because pbq=60, bp=bq

So bpq is equilateral.

So pq=bq

So cq:pq:pc=3:4:5

So PC square = pq square + cq squared.

So PQC is a right-angled triangle.