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As shown in the figure, in △ ABC, ∠ 1 + ∠ 2 = 230 ° and ad bisect ∠ BAC
∵ 1, ∵ 2 is the external angle of ABC,
∴∠1=∠BAC+∠ACB①；
∠2=∠BAC+∠ABC②，
∵∠1+∠2=230°，
ν ① + ②, 2 ∠ BAC + ∠ ACB + ∠ ABC = 230 ° ③,
∵∠BAC+∠ACB+∠ABC=180°④，
ν (4) - 3, BAC = 230 ° - 180 ° = 50 °,
∵ ad bisection ∵ BAC,
∴∠DAC=1
2∠BAC=1
2×50°=25°．
A: the degree of the DAC is 25 degrees
In △ ABC, ad is the center line on the edge of BC, e is a point on AC, be and ad intersect F, if ∠ FAE = AFE
It is proved that CG ∥ be is made, ad is delivered to g, and BG is connected
∵CG‖BE
∴∠BFD=∠CGD,∠FBD=∠GCD
And BD = CD
∴△BFD≌△CGD
BF=CG
∵CG‖BE
∴∠AFE=∠AGC
And ∠ AFE = ∠ FAE
∴∠AGC=∠FAE
∴AC=GC
Proved BF = CG
∴AC = BF
As shown in the figure, in the triangle ABC, ab = AC, BC = BD, ad = de = EB, calculate the degree of ∠ a
Let ∠ EDB be x degree ∠ AED = ∠ a = 2x degree ∠ EDC = 4x, so ∠ BDC = 3x, so ∠ C = 3x, so ∠ ABC = 3x3x + 3x + 2x = 180 degree, 8x = 180 degree, 2x = 45 degree, so ∠ a = 45 degree. If you don't understand this question, you can ask it again. If you are satisfied, please click "choose as satisfied answer". If you have other questions, please accept this question and send it separately
As shown in the figure, △ ABC, ad is the angular bisector, e and F are the points on AC and ab respectively, and ∠ AED + ∠ AFD = 180 °. What is the relationship between de and DF, and explain the reasons
The reason is: the over D is DM ? AB is in M, DN ? AC is in N, ∵ ad bisdivide ∵ BAC, ᙨ DM = DN, ∵ FMD = ∠ end = 90 ° and ∵ AED + ∵ AFD = 180 ᚜ D ⊸ ab ⊙ m, DN ? AC in N, ? ad bisdivide 9 BAC, 9 DM = DN, 9 FMD, 9 FMD, Δend
It is known that, as shown in the figure, in △ ABC, ad and AE are the bisectors of the height and angle of △ ABC respectively, if ∠ B = 30 ° and ∠ C = 50 ° (1) Find the degree of ∠ DAE; (2) What is the relationship between ∠ DAE and ∠ C - ∠ B? (no need to prove)
（1）∵∠B=30°，∠C=50°，
∴∠BAC=180°-30°-50°=100°．
∵ AE is the bisector of ∵ BAC,
∴∠BAE=50°．
In RT △ abd, ∠ bad = 90 ° - ∠ B = 60 °,
∴∠DAE=∠BAD-∠BAE=60°-50=10°；
（2）∠C-∠B=2∠DAE．
As shown in the figure, in △ ABC, O is the intersection point of high AD and be. Observe the graph and try to guess what kind of quantitative relationship exists between ∠ C and ∠ doe, and prove your conjecture conclusion
∠C+∠DOE=180°．
∵ ad, be is the height of ∵ ABC (known),
Ψ AEO = ∠ ADC = 90 ° (high meaning),
∵ DOE is the external angle of AOE (the concept of triangle external angle),
Ψ DOE = ∠ OAE + ∠ AEO (one outer angle of a triangle is equal to the sum of two nonadjacent interior angles)
=∠OAE+90°（∠AEO=90°）
=∠OAE+∠ADC（∠ADC=90°）
∴∠C+∠DOE=∠OAE+∠C+∠ADC=90°+90°=180°．
Another method: in the quadrilateral ceod, ∠ C + ∠ EOD + 90 ° + 90 ° = 360 °,
Then ∠ C + ∠ EOD = 180 °
In the triangle ABC, ad is the height on BC, CE is the upper midline of AB, DC = be, DG is perpendicular to CE, and G is the perpendicular foot Verification: 1. G is the midpoint of CE; 2. Angle B = 2 angle BCE
It is proved that e is the midpoint of AB in the right triangle ADB, so de = be = DC, so the triangle Dec is isosceles triangle, and DG is perpendicular to CE, so G is the midpoint of CE
Angle B = angle BDE = 2 angle BCE (external angle of triangle Dec, de = DC isosceles and equal angles)
Make up a picture by yourself
As shown in the figure, it is known that in △ ABC, ad is high, CE is midline, DC = be, DG ⊥ CE, G is perpendicular foot It is proved that: (1) g is the midpoint of CE; (2) B = 2 ∠ BCE
It is proved that: (1) connect De;
∵ ad ⊥ BC, e is the midpoint of ab,
﹤ De is the center line on the oblique side of RT △ abd, i.e., de = be = 1
2AB；
∴DC=DE=BE；
And ∵ DG = DG,
∴Rt△EDG≌Rt△CDG；（HL）
∴GE=CG，
/ / G is the midpoint of CE
(2) According to (1), be = de = CD;
∴∠B=∠BDE，∠DEC=∠DCE；
∴∠B=∠BDE=2∠BCE．
As shown in the figure, in the triangle ABC, ad, be and CF are the three midlines, which intersect at the same point g. are triangle AGF and triangle age equal? Why?
Because ad is the center line of the triangle ABC, the area of the triangle abd = the area of the triangle ACD. Similarly, the area of the triangle bgd = the area of the triangle CGD, so the area of the triangle ABG = the area of the triangle ACG
If the triangle DEF is a regular triangle, ad = be = cf. whether the triangle ABC is an equilateral triangle, if the proof process is given,
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