It is known that the triangle ABC, CA = CB, point O is on the vertical bisector of Ca and CB, and m.n is on a straight line AC.BC Go ahead, It is proved that CN + Mn = am

It is known that the triangle ABC, CA = CB, point O is on the vertical bisector of Ca and CB, and m.n is on a straight line AC.BC Go ahead, It is proved that CN + Mn = am

(1) Connect OC, intercept AQ = CN on am, connect OQ, ∵ o is the intersection point of vertical bisector of Ca and CB,

As shown in the figure, in △ ABC, ∠ cab = ∠ CBA = 45 °, CA = CB, point E is the midpoint of BC, CN ⊥ AE intersects AB with N and connects with en. Verification: AE = CN + en

It is proved that: extend CN to F, make CF = AE, connect BF,
∵∠CAB=∠CBA=45°，
∴∠ACB=90°，
∵CN⊥AE，
∴∠COE=90°，
∴∠CEA+∠1=90°，∠CEA+∠2=90°，
∴∠1=∠2，
In △ CAE and △ BCF
AC=CB
∠1=∠2
AE=CF
∴△CAE≌△BCF（SAS），
∴∠ACE=∠CBF=90°，CE=BF，
∵∠CBA=45°，
∴∠FBN=45°=∠EBN，
∵ e is the midpoint of BC,
∴CE=BE=BF，
In △ EBN and △ FBN
BE=BF
∠EBN=∠FBN
BN=BN
∴△EBN≌△FBN（SAS），
∴NE=NF，
∴AE=CN+EN．

As shown in the figure, ∠ ACB = 90 ° CA = CB in △ ABC, point D is a point outside the shape of △ ABC, and point D is on the vertical bisector of AC. if ∠ BCD = 30 °, calculate the value of ∠ abd

In ∵ ABC, ∵ ACB = 90 °, CA = CB,
ν Δ ABC is an isosceles right triangle,
∴∠ACB=90°，∠CAB=∠CBA=45°，
∵∠BCD=30°，
∴∠ACD=60°，
∵ D on the vertical bisector of AC,
∴CD=AD，
The △ ACD is an equilateral triangle,
∴AC=CD=AD，
∴DC=AC=BC，
∴∠CBD=∠CDB=75°，
∴∠ABD=∠CBD-∠CBA=75°-45°=30°．

It is known that in △ ABC, CA = CB, the intersection point O of the vertical bisector of Ca and CB is on AB, m and N are on the straight line AC and BC, respectively, ∠ mon = ∠ a = 45 °

In general, the question is: to prove: CN + Mn = am or the relationship between CN, Mn and am
Verification method: connect OC, intercept AQ = CN on am, connect OQ,
∵ o is the intersection point of the vertical bisector of Ca and CB, ᙽ OC = OA = ob,
∵ AC = BC, ᙽ OC ⊥ AB, CO bisection ∵ ACB,
Ψ a = ∠ B = 45 ° i.e. ∠ ACB = 90 °,
Ψ OCN = 45 °, i.e. ∠ OCN = ∠ a = 45 °,
In △ AOQ and △ con,
AQ=CN,∠A=∠OCN,OA=OC,
∴△AOQ≌△CON,
∴OQ=ON,∠AOQ=CON,
∵OC⊥AB,
∴∠AOC=∠AOQ+∠COQ=90°,
Ψ con + ∠ CoQ = 90 °, i.e. ∠ QON = 90 °,
And ∠ mon = 45 ° and ∠ Qom = 45 °,
In △ Qom and △ NOM,
OQ=ON,∠MON=∠QOM,OM=OM,
∴△QOM≌△NOM,
∴QM=NM,
Then am = AQ + QM = CN + Mn;
Hope can help you, hope to adopt

It is known that, as shown in the figure, ∠ ACB = 90 ° D, e are two points on AB, and ad = AC, be = BC

∠DCA=(180-∠A)÷2
=90-0.5∠A
∠BCE=(180-∠B)÷2
=90-0.5∠B
∠DCA+∠BCE=90-0.5∠A+90-0.5∠B
=180-0.5(∠A+∠B)
=180-0.5×90
=135
∠DCA+∠BCE-∠ACB=∠DCE=45°

As shown in the figure, in the isosceles right triangle ABC, AC = BC, ∠ ACB = 90 °, D and E are two points on AB, and ad = 6, be = 8, and ∠ DCE = 45 °, then the length of De is () A. 14 B. 9 C. 10 D. 11

Then △ ADC ≌ △ BCF,  BF = ad = 6, ∠ CBF = ∠ a = 45 °, EBF = ∠ ABC + ∠ CBF = 90 °, in right angle △ bef, EF = be2 + BF2 = 62 + 82 = 10, ∵ ACB = 90 °, DCE = 45 °, and ? 1 =

It is known that in the right triangle ABC, ∠ ACB = 90 °, ad = AC, be = BC, we can find the degree of ∠ DCE Question supplement: ∠ A is the top angle, a connection C ∠ C is a right angle, C connection B is the bottom line, a connection B is a diagonal line of the triangle, AB straight line respectively has e connection C, D connection C. (because can't draw graph here) thank you for your help!

∠ a + ∠ DCE + ∠ ADC = 180 degrees
∠ B + ∠ ECB + ∠ BEC = 180 degrees
A + ∠ B + 2 ∠ BEC + 2 ∠ ADC = 360 degrees
∠ a + ∠ B + ∠ C = 180 degrees
2 ∠ BEC + 2 ∠ ADC - ∠ C = 180 degrees
2 ∠ BEC + 2 ∠ ADC = 270 degrees
∠ bed + ∠ ADC = 135 degrees
∠ DCE = 45 degrees

In △ ABC, ∠ ACB = 90 °, points D and E are on AB, and ad = AC, BC = be. Find the degree of ∠ DCE

∵AD=AC，BC=BE，
∴∠ACD=∠ADC，∠BCE=∠BEC，
∴∠ACD=（180°-∠A）÷2①，∠BCE=（180°-∠B）÷2②，
∵∠A+∠B=90°，
ν ① + ② - ∠ DCE, ∠ ACD + ∠ BCE - ∠ DCE = 180 ° - (∠ a + ∠ b) ÷ 2 - ∠ DCE = 180 ° - 45 ° - ∠ DCE = 135 ° - ∠ DCE = 90 °,
∴∠DCE=45°．

As shown in the figure, in RT △ ABC, ∠ ACB = 90 °, BD bisection ∠ ABC, CE vertical BD, find the degree of ∠ DCE

∵AC=AE,BC=BD
∴∠ACE=∠AEC,∠BCD=∠BDC
∵∠ACB=100°
∴∠ACE+∠BCD=∠AEC+∠BDC
=100°+∠DCE ①
∵∠AEC+∠BDC+∠DCE=180° ②
Replace ① with ②
∴100°+∠DCE+∠DCE=180°
It is found that ∠ DCE = 40 degrees

As shown in the figure, CD is the height on the hypotenuse ab of RT △ ABC, CE is the bisector of BCA, ∠ a = 32 ° and the degree of ∠ DCE is obtained

One
∵ a = 32 °, △ ABC is RT triangle
∴∠CBD=∠ACB-∠A=90°-32°=58°
∵ CD is the height on the hypotenuse ab of rtabc
∴∠DCB=∠CDB-∠B=90°-58°=32°
∵ CE is the bisector of ∵ BCA
∴∠BCE＝45°
The △ AEC is an isosceles triangle
∴∠A=∠ECA=32°
∴∠DCE＝∠BCE－∠BCD＝45°－32°＝13°