It is known that AOB 90 and OM are the angular bisector of AOB, and the following questions are answ

1 (1) The perpendicular intersection of P and E is right angles to the Perpendicular intersection of P and O and F of P and O.

And because CPD is at right angles, EPC is equal to FPD, and because PE is equal to PF, PEC and PFD are congruent.

So pc=pd

POD is similar to PDG ( PGD= CGO= POD+ CDO=45°+ CDO+CDO=45°+ CDOs because PDO= PDC+ CDO=45°+ CDOs so PGD= PDOs and PDC= Pods so PODs are similar to PDGs), so the ratio of the area of PODs to PDGs is 3:2

2) Extend the PC interchange ob reverse extension line with e so that OE=OD (because OE=OD, OC=OC, OC is perpendicular so the triangle CEO congruent triangle CEO i.e. angle CDO=angle CEO and because the angle EPD is 90°, so the triangle ped congruent triangle CDO) So o is the midpoint of DE Let m be the midpoint of PO then Mo is perpendicular to PD because pm=md so triangle pom congruence with triangle mod So op=od=1

It's been a long time since I've done this kind of question... It should be done right ... If you still don't understand something, you can ask me again.

1.(1) If you connect cd, then o, d, p, c are contoured, so pcd= pod pdc= cop om is the angular bisector of aob, then pod = cop, then pcd= pdc so pc=pd

2) This question is a bit wrong, PG cannot be equal to PD

Answer: Because om and on are the angular bisectors of AOC and BOC respectively: com= aoc 2

con=∠boc/2

Subtract the two formulas to obtain:

com- con=( aoc- boc) 2So: mon= aob 2=120° 2=60°So: mon=60°

1 (1) The perpendicular intersection of P and E is right angles to the perpendicular intersection of P and F Peo and PFO.

And because CPD is at right angles, EPC is equal to FPD, and because PE is equal to PF, PEC and PFD are congruent.

So pc=pd

Pod is similar to PDG ( PGD= CG= POD= Pod+ CDO=45°+ CDO+PDO=45°+ CDOs so PGD= PDO= PDOs and PDC= Pods, so PODs are similar to PDGs), so the ratio of the area of PODs to PDGs is 3:2

2) Extend the PC interchange ob reverse extension line with e so that OE=OD (because OE=OD, OC=OC, OC is perpendicular to CEO, so the hole is open to the fiber triangle CEO, the congruent triangle CEO is the angle CDO=Angle CEO, and because the angle EPD is 90°, so the triangle is congruent triangle CDO) So O is the midpoint of DE, let M be the midpoint of PO, then Mo is perpendicular to PD because PM=MD, so the triangle POM congruence is with the triangle mod, so OP=OD=1

It's been a long time since I've done this kind of question... It should be done right ... If you still don't understand something, you can ask me again.

What about the figure? The first question should be related to "the distance from any point on the bisector of the angle to both sides of the angle is equal." ”

2∠aob=1

2×50°=25°，∠bon=1

2∠cob=1

2×80°=40°，∠mon=∠bon-∠aom=40°-25°=15°；

When the positional relationship between OA and BOC is shown in Figure 2, OM is the bisector of AOB and ON is the bisector of BOC, AOB=50°, COB=80°, BOM=1

2∠aob=1

2×50°=25°，∠bon=1

2∠boc=1

2 80°=40°, mon= BOM+ bon=25°+40°=65°, so choose c

<> solution: (1) om is the angular bisector of boc, on is the angular bisector of aoc, and aob=76°, 2 com+2 con=76°, mon=38°

3) From (1) and (2), when OC is found in any position in AOB, the value of Mon does not change, and when OC is outside AOB, the law does not hold