As shown in the figure, BD is the extension of the intersection of the middle line AE ⊥ BD on the edge AC of △ ABC at the point ECF ⊥ BD. be + BF. = 2bd is proved at the point F It's sloppy

As shown in the figure, BD is the extension of the intersection of the middle line AE ⊥ BD on the edge AC of △ ABC at the point ECF ⊥ BD. be + BF. = 2bd is proved at the point F It's sloppy

It's very simple! It's all the same!
∵AE⊥BD,CF⊥BD
Therefore, CFD = AED
Because BD is the middle line
So ad = CD
In △ AED and △ CFD
∠CFD=∠AED
AD=CD
∠ADE=∠CDF
So triangles are congruent
So Ed = DF
Because BD + BD = BF + DF + BD + de
So 2bd = be + BF

As shown in the figure, in RT △ ABC, ∠ C = 90 °, BD is the bisector of ∠ ABC, intersecting AC with D. if CD = 4, ab = 8, then the area of △ abd is______ ．

As shown in the figure, through point D, make de ⊥ AC at e, ∵ BD is the bisector of ∠ ABC, ∵ C = 90 °, ∵ de = CD = 4, ∵ abd area = 12ab, de = 12 × 8 × 4 = 16

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, intersecting BC at point D. if CD = 4m, ab = 10m, then the area of △ abd is______ m2．

When passing through point D, make de ⊥ AB at e, ∵ ∠ C = 90 °, ad is the bisector of ∠ BAC, ∫ de = CD = 4m ∫ s △ abd = 12 × 4 × 10 = 20 (M2)

As shown in the figure, in RT △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, intersecting BC at point D. if CD = 4m, ab = 10m, then the area of △ abd is______ m2．

When passing through point D, make de ⊥ AB at e, ∵ ∠ C = 90 °, ad is the bisector of ∠ BAC, ∫ de = CD = 4m ∫ s △ abd = 12 × 4 × 10 = 20 (M2)

As shown in the figure, it is known that ∠ AOB = 80 °, COD = 40 °, OM bisects ∠ BOD, on bisects ∠ AOC, (1) rotate ∠ COD in figure ① anticlockwise around o point, so that OC and

1) Because ∠ AOB = 90 °, AOC = 30 °, so ∠ BOC = 120 °. Because om bisects ∠ BOC, so ∠ com = 12 ∠ BOC = 60 °. Because on bisects ∠ AOC, so ∠ con = 1 / 2 ∠ a OC = 1 / 2 × 30 ° = 15 °. So ∠ mon = ∠ com - ∠ con = 60 ° - 15 ° = 45 °;
(2) When ∠ AOB = α and other conditions remain unchanged, imitating (1), we can get ∠ mon = 1 / 2 α (i.e. (a + ∠ AOC) △ 2 - ∠ AOC △ 2 = 1 / 2 α)
(3) By imitating (1), we can get ∠ mon = ∠ com - ∠ con

As shown in the figure, OM is the bisector of ∠ AOB, the ray OC is inside the ∠ BOM, and on is the bisector of ∠ BOC. It is known that ∠ AOC = 80 °, then the size of ∠ mon is equal to______ °．

∵ on bisection ∵ BOC ∵ con = ∵ Bon let ∵ con = ∵ Bon = x, ∵ MOC = y, then ∵ mob = ∵ MOC + ∵ BOC = 2x + y and ∵ om bisection ∵ AOB ∵ AOM = ∵ BOM = 2x + y ∵ AOC = ∵ AOM + ∵ MOC = 2x + y + y = 2 (x + y) ∵ AOC = 80 °∵ 2 (x + y) = 80 °∵ x + y = 40 ∵ mon = ∵ MOC + ∵ NOC = x + y = 40 ° so the answer is 40 °

When the angle AOB is 136 ° and the ray OC is inside the angle AOB, OM bisects the angle AOC and on bisects the angle BOC, the degree of the angle mon is

∵ om bisection ∠ AOC
∴∠COM＝∠AOC/2
∵ on bisection ∠ BOC
∴∠CON＝∠BOC/2
∴∠MON＝∠COM+∠CON＝（∠AOC+∠BOC）/2＝∠AOB/2＝136/2＝68°
I hope the building owner will adopt mine_

It is known that ∠ AOB = 90 °, OC is a ray, OM and on divide ∠ BOC and ∠ AOC equally, and calculate the value of ∠ mon

When OC is inside ∠ AOB, it is easy to get ∠ mon = 45 ° and when OC is outside ∠ AOB, it is necessary to discuss in four cases: acute angle, right angle, obtuse angle and flat angle. The results are 45 °, 45 °, 135 ° and 135 ° respectively. The conclusion is: the ∠ mon is 45 ° or 135 ° respectively

It is known that ∠ AOB = 136 °, ray OC is in the interior of ∠ AOB, OM bisects ∠ AOC, on bisects ∠ BOC, then ∠ mon is () °

∠MON=∠MOC+∠NOC=1/2∠AOC+1/2∠BOC=1/2（∠AOC+∠BOC）=1/2∠AOB=68°

It is known that ∠ AOB = 90 °, OC is a ray, OM and on divide ∠ BOC and ∠ AOC equally, and calculate the value of ∠ mon

When OC is inside ∠ AOB, it is easy to get ∠ mon = 45 ° and when OC is outside ∠ AOB, it is necessary to discuss in four cases: acute angle, right angle, obtuse angle and flat angle. The results are 45 °, 45 °, 135 ° and 135 ° respectively. The conclusion is: the ∠ mon is 45 ° or 135 ° respectively