As shown in the figure, point P is in ∠ AOB, and points m and N are the symmetrical points Mn of point P about OA and ob intersecting OA and ob at points E and F. if the perimeter of △ PEF is 15, the length of line segment Mn can be obtained

As shown in the figure, point P is in ∠ AOB, and points m and N are the symmetrical points Mn of point P about OA and ob intersecting OA and ob at points E and F. if the perimeter of △ PEF is 15, the length of line segment Mn can be obtained

According to the property of axisymmetry, EP = em, PF = FN, so the length of segment Mn = △ PEF,
Ψ Mn = me + EF + FN = PE + EF + pf = △ PEF perimeter,
∴MN=15cm．

As shown in the figure, P is in ∠ AOB, and points m and N are the symmetrical points of point P about AO and Bo respectively, and the intersection points E and F with AO and Bo. If the perimeter of △ PEF is 15, the length of Mn is calculated

The ∵ point m is the symmetric point of point P with respect to Ao, ∵ Ao bisects MP vertically, ∵ EP = em. similarly, PF = FN. ∵ Mn = me + EF + FN, ∵ Mn = EP + EF + PF, the perimeter of ∵ PEF is 15, ∵ Mn = EP + ef + pf = 15

As shown in the figure, P is in ∠ AOB, and points m and N are the symmetrical points of point P about AO and Bo respectively, and the intersection points E and F with AO and Bo. If the perimeter of △ PEF is 15, the length of Mn is calculated

The ∵ point m is the symmetric point of point P with respect to Ao, ∵ Ao bisects MP vertically, ∵ EP = em. similarly, PF = FN. ∵ Mn = me + EF + FN, ∵ Mn = EP + EF + PF, the perimeter of ∵ PEF is 15, ∵ Mn = EP + ef + pf = 15

As shown in the figure, P is in ∠ AOB, and points m and N are the symmetrical points of point P about AO and Bo respectively, and the intersection points E and F with AO and Bo. If the perimeter of △ PEF is 15, the length of Mn is calculated

The ∵ point m is the symmetric point of point P with respect to Ao, ∵ Ao bisects MP vertically, ∵ EP = em. similarly, PF = FN. ∵ Mn = me + EF + FN, ∵ Mn = EP + EF + PF, the perimeter of ∵ PEF is 15, ∵ Mn = EP + ef + pf = 15

As shown in the figure, OA is perpendicular to ob, OC is inside ∠ AOB, OM and on divide ∠ BOC and ∠ AOC equally

∠MON=∠MOC+∠NOC=1/2∠BOC+1/2∠AOC=1/2(∠BOC+∠AOC)=1/2∠AOB=1/2*90=45

As shown in the figure, it is known that OM and on divide ∠ AOC and ∠ BOC equally. If ∠ mon = 45 °, then OA ⊥ ob. Can you explain why?

∵ om bisection ∠ AOC
∴∠COM=∠AOC/2
∵ on bisection ∠ BOC
∴∠BON=∠CON=∠BOC/2
∵∠MON=∠COM-∠CON
=∠AOC/2-∠BOC/2
=(∠AOC-∠BOC)/2
=∠AOB/2=45
∴∠AOB=90
∴OA⊥OB

As shown in the figure, OA ⊥ ob, on bisects ∠ AOC, OM bisects ∠ BOC, and calculates the degree of ∠ mon
Big brother and big sister, help!

However, in either case, ∠ mon is 45 ° (1) if OC is in AOB, then on bisects ∠ AOC, set ∠ AON as ∠ 1, OM bisects ∠ BOC, and set ∠ BOM as ∠ 2, so in ∠ AOB, 2 ∠ 1 + 2 ∠ 2 =

As shown in the figure, OM and on are divided equally

∠AOM=∠MOC,∠CON=∠NOB
Therefore, AOM + nob = MOC + con = mon = 45 degrees
Therefore, AOB = AOM + nob + MOC + con = 2 mon = 90 degrees

When the angle AOB is known, the ray OC, OM bisecting angle AOC, on bisecting angle BOC are made through the point O. when OC is inside the angle AOB, it is proved that mon = half angle AOB

It is proved that when OC is in angle AOB, because om bisects angle AOC, MOC is equal to half AOC, because on bisects BOC, NOC is equal to half BOC, angle mon = MOC + NOC = 1 / 2aoc + 1 / 2boc, angle AOB = AOC + BOC
So it is proved that the angle mon = half angle AOB

The OC is different from OA and ob. If OM is known to bisect ∠ BOC and on is known to bisect ∠ AOC, then ∠ mon=

If OC is within the 90 ° angle of ∠ AOB, ∠ mon is half of 90 ° which is 45 °
If OC is in the range of 270 ° outside of ∠ AOB, ∠ mon is half of 270 ° which is 135 °