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In △ ABC, ∠ a = 1 2∠B=1 3 ∠ ACB, CD is the height of △ ABC, CE is the angular bisector of ∠ ACB, and the degree of ∠ DCE is calculated
A = 12, B = 13, ACB, ACB, ACB, ACB, ACB, ACB, ACB, ACB = 3, a, a, ACB = 3, a, a, a, a, a, a, a, a, a, a, a, B, a, a, a, a, a, B, a, a, a, a, B, a, B, a, B, a, a, B, a, B, a, a, B, a, B, a, B, a, B, a, B, a, a, B, a, a, B, a, B, a, B, a, B, a, B, a, C, a, a, C, a, a, C, a, C, a, C, a, C, a, a, a, C, a, a, C, a, × 90? =
In △ ABC, ∠ a = 1 2∠B=1 3 ∠ ACB, CD is the height of △ ABC, CE is the angular bisector of ∠ ACB, and the degree of ∠ DCE is calculated
∵∠A=1
2∠B=1
3∠ACB，
∴∠B=2∠A，∠ACB=3∠A，
∵∠A+∠B+∠ACB=180°，
∴∠A+2∠A+3∠A=180°，
The solution is ∠ a = 30 °,
∴∠ACB=90°，
∵ CD is the height of ∵ ABC,
∴∠ACD=90°-30°=60°，
∵ CE is the angular bisector of ∵ ACB,
∴∠ACE=1
2×90°=45°，
∴∠DCE=∠ACD-∠ACE=60°-45°=15°．
As shown in the figure, in the known triangle ABC, ∠ ACB = 90 °, CD, CE, trisection ∠ ACB, and CD ⊥ AB, please explain: ab = 2BC; CE = AE = be
To solve AB = 2BC, the reasons are as follows: ∵ CD, CE trisection ACB (known)
As shown in the figure, we know that in △ ABC, ∠ ACB = 90 ° CD, CE are trisection ∠ ACB, and CD ⊥ AB, please prove that: (1) ce is the central line of RT △ ABC （2）AB=2BC
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As shown in the figure, in △ ABC, ∠ C is a right angle. The high Cd on AB and the central line CE exactly divide ∠ ACB. If AB = 20, what are the two acute angles of △ ABC and the values of AD, de and EB?
C is a right angle, CD and CE exactly divide ACB into three equal parts, ACD = ∠ DCE = ∠ ECB = 13 × 90 ° = 30 °, CD is high, a = 90 ° - ∠ ACD = 90 ° - 30 ° = 60 °,
As shown in the figure, in △ ABC, ∠ C is a right angle. The high Cd on AB and the central line CE exactly divide ∠ ACB. If AB = 20, what are the two acute angles of △ ABC and the values of AD, de and EB?
C is a right angle, CD and CE exactly divide ACB into three equal parts, ACD = ∠ DCE = ∠ ECB = 13 × 90 ° = 30 °, CD is high, a = 90 ° - ∠ ACD = 90 ° - 30 ° = 60 °,
It is known that, as shown in the figure, the triangle ABC and the triangle ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90 °, D is a point on the edge of AB, and it is proved that BD = AE
Tuna?
It is known that △ ABC and △ ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90 ° and D is a point on AB edge It is known that △ ABC and △ ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90 ° and D is a point on AB edge Confirmation: BD = AE
Proof: because triangle ABC and triangle ECD are isosceles right triangle
So EC = CD, ab = BC, angle ACB = angle DCE = 90 degrees
Because angle ECD = angle ECA + angle ACD, angle ACB = angle ECB + angle ACD
So angle ECA = angle DCB
So the triangle ECA is equal to the triangle DCB
So DB = AE
As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC
It is proved that DN ⊥ AC, DM ⊥ AB are made by D,
△ the area of DBF is: 1
2BF•DM，
The area of △ DCE is 1
2DN•CE，
The area of ∵ △ DCE and △ DBF are equal,
∴1
2BF•DM=1
2DN•CE，
∵CE=BF，
∴DM=DN，
Ψ ad bisection ∠ BAC (points with equal distance to both sides of the corner are on the bisector of the angle)
D. E, f are the points on the three sides of △ ABC, CE = BF, the areas of △ DCE and △ DBF are equal In △ ABC, BD = DC, ED, vertical DF: be + CF > EF
(1) Let d make AB, AC perpendicular lines intersect g, h respectively
Since CE = BF, △ CDE, △ BDF have the same area, then
CE*DH=BF*DG => DH=DG
In the vertical △ ADG and △ ADH
DG=DH
Ad coincidence
So △ ADG, △ ADH are congruent
So ∠ bad = ∠ CAD
(2)∠EDF=90°
So ∠ BDE, ∠ CDF > 90 °
So be > de, CF > DF
In the right angle △ def,
∠EDF=90°
So de + DF > EF
In addition, be > de, CF > DF
BE+CF>EF
RELATED INFORMATIONS
- 1. As shown in the figure, D, e and F are points on the three sides of △ ABC, and the areas of CE = BF, △ DCE and △ DBF are equal Verification: ad bisection ∠ BAC
- 2. As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle
- 3. As shown in the figure, in diamond ABCD, AE ⊥ BC, perpendicular foot is e, angle DAE = 2, angle BAE, BD = 15, find the length of AC, ab
- 4. As shown in the figure, in △ ABC, D is a point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
- 5. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
- 6. As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad
- 7. In the triangle ABC, the points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and the ad bisection angle BAE (1) when the angle BAC is equal to 90 degrees In the triangle ABC, points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and ad bisects the angle BAE (1) When the angle BAC is equal to 90 degrees, it is proved that BD is equal to AC (2) When the angle BAC is not equal to 90 degrees, is there still BD equal to AC? Explain the reason
- 8. As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De
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- 10. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 11. 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
- 12. As shown in the figure, in the RT triangle ABC, the angle c = 90 degrees, the DB bisector angle ABC intersects AC at point D, De is the vertical bisector of AB and intersects AB at point E (1) Find the degree of angle A (2) If BC = 6, AC = 8, then the circumference of the triangle BDC is
- 13. This is the geometry of grade two, which is related to parallelogram In the RT triangle ABC, the angle c = 90 degrees, M is the midpoint of AB, am = an, Mn is parallel to AC, and Mn = AC
- 14. As shown in the figure, ∠ ABC = ∠ DBE = 90 °, DB = be, ab = BC (1) Confirmation: ad = CE, ad ⊥ CE; (2) If the external conditions of △ DBE rotating around point B to △ ABC remain unchanged, then the conclusion in (1) still holds? Draw a graph to prove your conclusion
- 15. In the triangle ABC, BD = DC = AC, D is the midpoint of BC and E is the midpoint of DC. It is proved that the bisection angle BAE of AD
- 16. In △ ABC, ab = BC, ∠ B = 90 °, M is the midpoint of AC, D.E is the moving point on AB and BC respectively, and BD = CE, the shape of △ DEM is obtained and proved
- 17. As shown in the figure, in the isosceles triangle ABC, ab = AC, D is a point on the extension line of AB, BD = AB, CE is the center line on waist AB, and verify; CD = 2ce needs specific process
- 18. As shown in the figure, in RT △ ABC, ab = AC, ∠ a = 90 °, point D is any point of BC, DF ⊥ AB is in F, de ⊥ AC is in E, M is the midpoint of BC. Try to judge what shape △ MEF is and prove your conclusion
- 19. D is the middle of the edge BC of the triangle ABC, De is vertical AC, DF is vertical AB, the vertical feet are e, F, and BF = CE, judge the shape of triangle ABC Is it an isosceles right triangle? Give proof
- 20. As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained