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As shown in the figure, in △ ABC, ∠ ABC = ∠ C = 2 ∠ a, BD ⊥ AC intersects AC at point D, then ∠ DBC=______ .
∵∠ABC=∠C=2∠A，
ν let ∠ a = x (degree), then ∠ ABC = ∠ C = 2x,
∵∠ABC+∠C+∠A=180°，
∴2x+2x+x=5x，
That is, 5x = 180,
X = 36 ° is obtained,
∴∠C=72°，
∵BD⊥AC，
∴∠BDC=90°
∴∠DBC=90°-∠C=90°-72°=18°．
So the answer is: 18 degrees
As shown in the figure, given that AE is parallel to BC and ad is parallel to BC, then the angle DAE is 180 degrees. Why? D_______ A________ E / ) B ------ C
AE is parallel to BC, ad is parallel to BC
Then: D, a and E are collinear
So: angle DAE = 180 degrees
Known: as shown in the figure: angle 1 plus angle 2 equals 180 degrees, angle a equals angle c BC square angle DBE to verify ad bisection angle BDF
? 1 + ∠ 2 = 180 °, ab ∥ CD, (two lines are parallel to each other's external angle) ? a = C, BC bisection DBE, DBC = ∠ CBE = ∠ C = ∠ a = ∠ ADF, (two lines are parallel at the same angle) ? ad ∥ BC, (the same position angle is equal, two lines are parallel) a = ∠ C, BC is bisection ﹤ DBE, (equal exchange), ﹤ ad bisection ∫ BDF
As shown in the figure, given angle Abe = angle DEB = 180 ° angle 1 = angle 2, try to explain the reason why angle f = angle G
∵∠ABE+∠DEB=180°
/ / AC ∥ de
﹤ CBE = ∠ DEB (two lines are parallel, and the internal staggered angle is equal)
∵∠1=∠2∴∠CBE-∠1=∠DEB-∠2
Namely ∠ FBE = ∠ Geb BF ∥ Ge (the internal staggered angles are equal and the two lines are parallel)
Ψ f = ∠ g (two straight lines are parallel, and the internal stagger angle is equal)
Can you give me some points
As shown in the figure, it is known: ∠ Abe + ∠ DEB = 180 °, 1 = ∠ 2. Try to explain the relationship between ∠ F and ∠ g, and explain the reasons
The reasons are as follows:
∵∠ABE+∠DEB=180°，
∴AC∥DE，
∴∠CBE=∠DEB，
∵∠1=∠2，
∴∠FBE=∠GEB，
∴BF∥GE，
∴∠F=∠G．
As shown in the figure, it is known: ∠ Abe + ∠ DEB = 180 °, 1 = ∠ 2. Try to explain the relationship between ∠ F and ∠ g, and explain the reasons
The reasons are as follows:
∵∠ABE+∠DEB=180°，
∴AC∥DE，
∴∠CBE=∠DEB，
∵∠1=∠2，
∴∠FBE=∠GEB，
∴BF∥GE，
∴∠F=∠G．
As shown in the figure, given ∠ 1 + ∠ 2 = 180 ° and ∠ B = ∠ 3, can you judge the size relationship between ∠ ACB and ∠ AED? Give reasons
∠ AED = ∠ ACB. (2 points)
The reasons are as follows: ∵ 1 + ∠ 2 = 180 °, 1 + ∠ 4 = 180 °,
∴∠2=∠4，
∴BD∥FE
Ψ 3 = ∠ ade (4 points)
∵∠3=∠B，
∴∠B=∠ADE
▽ de ‖ BC, (8 points)
﹤ AED = ∠ ACB. (10 points)
As shown in the figure, given angle 1 + angle 2 = 180 degrees, angle B = angle 3, can you judge the relationship between angle ACB and angle AED? Please explain the reasons
(1) Equality
∵∠2=∠3+∠EDF
∠1+∠2=180°
∴∠1+∠3+∠EDF=180°
∵∠3=∠B
∴∠1+∠B+∠EDF=180°
That is ∠ EDB + ∠ B = 180 degrees
∴DE//BC
∴∠AED=∠C
As shown in the figure, BD and CE are the heights of △ ABC, and it is proved that: ∠ AED = ∠ ACB
It is proved that: ∵ ADB = ∠ AEC = 90 °, a = ∠ a,
∴△ABD∽△ACE．
∴AD
AE＝AB
AC．
And ∠ a = ∠ a,
∴△ADE∽△ABC．
∴∠AED=∠ACB．
As shown in the figure, BD and CE are the heights of △ ABC, and it is proved that: ∠ AED = ∠ ACB
It is proved that: ∵ ADB = ∠ AEC = 90 °, a = ∠ a,
∴△ABD∽△ACE．
∴AD
AE＝AB
AC．
And ∠ a = ∠ a,
∴△ADE∽△ABC．
∴∠AED=∠ACB．
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