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In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
∠ABC=180-∠CBD=180°90°=90°，
∠ACB=180°-∠A-∠ABC，
=180°-30°-90°，
=60°，
∠BCE=180°-∠ACB，
=180°-60°，
=120°．
A: BCE is 120 degrees
In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
∠ABC=180-∠CBD=180°90°=90°，
∠ACB=180°-∠A-∠ABC，
=180°-30°-90°，
=60°，
∠BCE=180°-∠ACB，
=180°-60°，
=120°．
A: BCE is 120 degrees
In the triangle ABC, CD is perpendicular to AB and D. if ad is equal to 2bd, AC is equal to 5, and BC is equal to 4, the length of BD is calculated
If BD = x, then ad = 2x, there are: 5 ^ 2 - (2x) ^ 2 = 4 ^ 2-x ^ 2
3x^2=9
x^2=3
x=√3
As shown in the figure, ∠ ABC = 60 ° in △ ABC, ad and CE are the heights on BC and ab respectively, and F is the midpoint of AC. try to judge the shape of △ def and prove your conclusion
Connected with EF, △ DEF is an equilateral triangle, with ∠ ABC = 60 °,
Easy to get: be
BC＝BD
AB＝1
2．
∴△BDE∽△BAC，
∴DE
AC＝BD
AB＝1
2，
∴DE=1
2AC．
And ∵ f is the midpoint,
In RT △ ADC, DF = 1
2Ac, in RT △ ace, EF = 1
2AC．
So de = DF = EF
That is, △ DEF is an equilateral triangle
In the triangle ABC, the angle a = 90 degrees, ad is perpendicular to BC and D. if AB = radical 6, CD = 1, then what is the area of the triangle ABC?
∵∠B+∠C=90°,∠DAC+∠C=90°∴∠B=∠DAC,∠BDA=∠ADC=90°
∴△ABD∽△CAD
∴AB\AC=AD\CD=DB\AD
ν AC × ad = ab × CD = root 6 × 1 = root 6
AD×AD=CD×BD=1×BD=BD
In RT △ abd: ab squared = ad squared + DB squared = ad squared + ad quartic
6 = the square of AD + the fourth power of ad
The square root of AD-2 is AD-2
ν AC × root 2 = root 6
/ / AC = radical 3
∴S△ADC=AB×AC÷2
=Root 6 × root 3 △ 2
=2
Isosceles triangle ABC, AB equals AC, angle c equals 30 degrees, AB vertical ad, ad equals 2, find the length of BC Why is BD 4 cm
If the angle B is equal to the angle c is equal to 30 degrees, ABC is a right triangle. According to the fact that the right angle to which 30 degrees are opposite is half of the hypotenuse, we can see that BD = 2ad, and because ad = 2, BD is equal to 2 × 2 = 4
Junior high school students, study hard,
As shown in the figure, ⊙ o radius is 2, chord BD = 2 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, the area of quadrilateral ABCD is calculated
Connect OA to BD at point F, connect ob,
∵ OA is on the diameter and point a is the midpoint of arc BD,
∴OA⊥BD，BF=DF=
Three
In RT △ BOF
From Pythagorean theorem, of2 = ob2-bf2 is obtained
OF=
22−(
3)2=1
∵OA=2
∴AF=1
∴S△ABD=2
3×1
2=
Three
∵ point E is the midpoint of AC
∴AE=CE
And ∵ ∵ ade and ∵ CDE are the same
∴S△CDE=S△ADE
∵AE=EC，
∴S△CBE=S△ABE．
∴S△BCD=S△CDE+S△CBE=S△ADE+S△ABE=S△ABD=
Three
﹤ s quadrilateral ABCD = s △ abd + s △ BCD = 2
3．
As shown in the figure, ⊙ o radius is 2, chord BD = 2 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, the area of quadrilateral ABCD is calculated
In RT △ BOF, from the Pythagorean theorem, of2 = ob2-bf2of = 22 − (3) 2 = 1 ∵ OA = 2 ∵ AF = 2 ∵ AF = 1 ? OA ∵ OA ∵ OA ∵ OA is on the diameter and point a is the midpoint of arc BD ? OA ⊙ OA ⊙ OA ∵ OA is on the diameter and point a is the midpoint of arc ? OA ? OA is on the diameter and point a is the midpoint of the arc ? AE = CE ? ade and a
As shown in the figure, ⊙ o radius is 2, chord BD = 2 3, a is the midpoint of arc BD, e is the midpoint of chord AC, and on BD, the area of quadrilateral ABCD is calculated
Connect OA to BD at point F, connect ob,
∵ OA is on the diameter and point a is the midpoint of arc BD,
∴OA⊥BD，BF=DF=
Three
In RT △ BOF
From Pythagorean theorem, of2 = ob2-bf2 is obtained
OF=
22−(
3)2=1
∵OA=2
∴AF=1
∴S△ABD=2
3×1
2=
Three
∵ point E is the midpoint of AC
∴AE=CE
And ∵ ∵ ade and ∵ CDE are the same
∴S△CDE=S△ADE
∵AE=EC，
∴S△CBE=S△ABE．
∴S△BCD=S△CDE+S△CBE=S△ADE+S△ABE=S△ABD=
Three
﹤ s quadrilateral ABCD = s △ abd + s △ BCD = 2
3．
As shown in the figure, the radius of circle O is 2, chord BD = 2, root sign 3, a is the midpoint of arc BD, and E is the focus of chord AC, and it is on BD
Because a is the midpoint of arc BD and AF is perpendicular to BD, then f is the midpoint of BD. because r = 2, BD = 2 √ 3, BF = √ 3, Bo = 2, of vertical BF, then of = 1, that is AF = 1, then the area of triangle abd = BD * AF / 2 = 2 √ 3 *
RELATED INFORMATIONS
- 1. It is known that AC and BD are two mutually perpendicular chords of the circle O: x2 + y2 = 4, and the perpendicular foot is m (1, 2) The maximum area of the quadrilateral ABCD is () A. 4 B. 4 Two C. 5 D. 5 Two
- 2. It is known that AB is the diameter of ⊙ O. the straight line BC and ⊙ o are tangent to point B. through a, ad ⊙ OC intersects ⊙ o at point D and connects CD Proof: CD is tangent of ⊙ o
- 3. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
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- 5. As shown in the graph, points D, e and F are the points on the edges BC, Ca and ab of the triangle ABC, De is parallel to Ba, DF is parallel to Ca, and it is proved that ∠ FDE = ∠ a
- 6. As shown in the figure, the radius of ⊙ o is known to be 1, De is the diameter of ⊙ o, the tangent of ⊙ o is ad, C is the midpoint of AD, AE intersects ⊙ o at point B, and the quadrilateral bcoe is a parallelogram (1) Find the length of AD; (2) Is BC tangent to ⊙ o? If so, give proof; if not, give reasons
- 7. 0
- 8. AB is the diameter of the circle, the radius OC ⊥ AB, the chord EF is parallel AB through the midpoint D of OC. It is proved that the angle Abe is equal to 15 ° and the angle CBE is equal to 30 °
- 9. In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship
- 10. Two equal line segments AB and CD coincide in one third. M and N are the midpoint of AB and CD respectively. If Mn = 12cm, find the length of ab
- 11. 0
- 12. 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
- 13. 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
- 14. 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
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- 16. 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
- 17. 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
- 18. 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
- 19. 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
- 20. 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