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Ad is the height of triangle ABC, be is the midline angle on the side of AC, CBE = 30 degrees to prove ad = be
When passing through point E, EF ⊥ BC is f,
There are: EF ‖ ad
Because EF = be · sin ∠ CBE = (1 / 2) be,
So be = 2ef = ad. and because EF ‖ ad, AE = EC,
So EF is the median of △ ACD,
It can be concluded that ad = 2ef = be
In the triangle ABC, ad is the height of BC, be is the middle line of AC, and the angle CBE is equal to 30 degrees
If we make ef ⊥ BC through point e to F, then we have EF ∥ ad
Because EF = be · sin ∠ CBE = (1 / 2) be,
So be = 2ef = ad
Because, EF ‖ ad, AE = EC,
So EF is the median of △ ACD,
It can be concluded that ad = 2ef = be
As shown in the figure, in △ ABC, ad is high, be is the middle line, ∠ CBE = 30 °. Verification: ad = be
Extend CB to f so that CB = BF, connect AF,
Be is the median line of AF, that is 2be = AF,
Then AF / / be,
And the angle AFD = 30 degrees,
So AF = 2ad,
So be = ad
As shown in the figure, AB is the diameter of ⊙ o, the tangent of ⊙ o is made through point B, the extension line of cross chord AE is made at point C, OD ⊥ AC is made, and the perpendicular foot is d. if ∠ ACB = 60 ° BC = 4, the length of de is______ ．
∵ BC cuts ⊙ o to B, ∵ ABC = 90 °, ∵ ACB = 60 °, ∵ BAC = 30 °, ∵ AC = 2BC = 8. According to Pythagorean theorem, ab = ac2 − BC2 = 43, ∵ OA = 12ab = 23, ∵ OD ⊥ AE, ∵ ADO = 90 °, ∵ od = 12oa = 3. In △ ADO, according to Pythagorean theorem, ad = 3, ∵ OD ⊥ AE, OD over center O, ∵ ad = de = 3, (vertical diameter theorem), so the answer is: 3
As shown in the figure, AB is the diameter of ⊙ o, points D and E are the trisection points of the semicircle, and the extension lines of AE and BD intersect at point C. If CE = 2, the area of the shadow part in the figure is ()
A. 43π-3B. 23πC. 23π-3D. 13π
Connecting OE and OD, the points D and E are triangles of semicircle, and the area of shadow in the figure is s-sector oae-s △ OAE + s-sector ode = 60 ·π· 22
As shown in the figure, AB is the diameter of ⊙ o, points D and E are the trisection points of the semicircle, and the extension lines of AE and BD intersect at point C. If CE = 2, the area of the shadow part in the figure is ()
A. 43π-3B. 23πC. 23π-3D. 13π
Connecting OE and OD, the points D and E are triangles of semicircle, and the area of shadow in the figure is s-sector oae-s △ OAE + s-sector ode = 60 ·π· 22
If AB is the diameter of circle O and D is the midpoint of BC, then the relationship between AB and AC is obtained,
AB=AC
Connect ad,
∵ AB is the diameter, ∵ ad ⊥ BC
∵ D is the midpoint of BC, ∵ BD = DC
Easy syndrome △ ADB ≌ △ ADC (SAS)
∴AB=AC
As shown in the figure, BC is the diameter of ⊙ o, a is the point on ⊙ o, cross the extension line of BA at point D, take the midpoint e of CD, the extension line of AE and the extension line of BC at point P. (1) Description: AP is the tangent line of ⊙ o; (2) if OC = CP, ab = 6, find the length of CD
(1) It is proved that: connecting Ao, AC (as shown in the figure). ∵ BC is the diameter of ⊙ o, ∵ BAC = ∠ CAD = 90 °. ∵ e is the midpoint of CD, ∵ CE = de = AE. ∵ ECA = ∠ EAC. ∵ OA = OC, ∵ OAC = ∠ OCA. ∵ CD is the tangent line of ⊙ o, ∵ CD ⊥ OC. ∵ ECA + ∠ OCA = 90 °. ∵ EAC + ∠ OAC = 90 °. OA ⊥ AP. ∵ A is the point on ⊙ o, and ⊙ AP is the tangent line of ⊙ o; (2) In the RT △ OAP, the \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= 23cos30 ° = 2332 = 4
As shown in the figure, in rectangular ABCD, AE ⊥ BD is in E, ∠ DAE = 3 ∠ BAE, calculate the degree of ∠ BAE, ∠ EBC
Because ∠ DAE = 3 ∠ BAE, because the quadrilateral ABCD is rectangular, so ∠ bad = 90, so ∠ BAE = 22.5
Because AE is perpendicular to BC, so ∠ abd = 67.5, because rectangle, so ∠ ABC = 90, so ∠ EBC = 22.5
So, BAE = 22.5 and EBC = 22.5
In rectangle ABCD, AE is vertical BD, angle DAE: angle BAE = 3:1, calculate the degree of angle BAE and angle EAO
∠BAE=22.5°
∠EAO=45°(∠DAO=∠BAE=22.5°)
RELATED INFORMATIONS
- 1. In rectangular ABCD, AE is perpendicular to BD, and the angle DAE = 2 angle BAE?
- 2. As shown in the figure, in RT △ ABC, ∠ C = 90 °, am is the middle line on the side of BC, sin ∠ cam = 35, then the value of tanb is () A. 32B. 23C. 56D. 43
- 3. As shown in the figure, in the isosceles trapezoid ABCD, it is known that ad ∥ BC, ab = CD, AE ⊥ BC are in E, ∠ B = 60 °, DAC = 45 ° and AC = 6?
- 4. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 108 °, D is on AC and BC = AB + CD, the proof is: BD bisects ∠ ABC
- 5. In a convex quadrilateral ABCD, Da = DB = DC = BC, then the degree of the largest angle in the quadrilateral is () A. 120°B. 135°C. 150°D. 165°
- 6. In RT △ ACB, ∠ ACB = 90 °, AC = 3, BC = 4, the moving circle O passing through point C is tangent to the hypotenuse AB at the moving point P, and the value range and maximum value of CP are obtained
- 7. In △ ABC, ∠ C = 90 °, AC = 4cm, BC = 5cm, point D is on BC, and CD = 3cm. There are two moving points P and Q starting from point a and point B at the same time, where point P moves along AC to terminal C at the speed of 1cm / s, and point Q moves along BC to terminal C at the speed of 1.25cm/s. Passing point P, PE ‖ BC is connected to point E, and EQ is connected (1) The length of AE and De is expressed by the algebraic expression containing x; (2) when the point Q moves on BD (excluding points B and D), let the area of △ EDQ be y (cm2), find the functional relationship between Y and time x, and write out the value range of the independent variable x; (3) when the value of X is, the △ EDQ is a right angle triangle
- 8. As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
- 9. As shown in the figure, given that point E is on the edge ab of △ ABC, point D is on the extension line of Ca, and point F is on the extension line of BC, what is the size relationship between ∠ ACF and ∠ D? Please give reasons
- 10. As shown in the figure, the triangle ABC is an equilateral triangle, and D is the midpoint on the side ab. it is known that the area of the triangle BDE is 5 square centimeters. Find the area of the equilateral triangle ABC
- 11. As shown in Figure 7-11, in the known triangle ABC, ad is perpendicular to D, BC is equal to D, AE is equal to ∠ BAC (2), if ∠ B ＞ C, try to explain ∠ DAE = 1 / 2 (∠ B - ∠ C)
- 12. It is known that in the oblique triangle ABC, the straight line with high BD and CE intersects at h, and the angle a = 45 degrees, so the degree of the angle BHC can be calculated? No picture, if you can,
- 13. In known quadrilateral ABCD, angle abd = 12 °, angle DBC = 36 °, angle ACB = 48 ° and angle ACD = 24 °, try to find the degree of angle ADB I can draw pictures according to the title,
- 14. As shown in the figure, in △ ABC, be bisects ∠ ABC intersects AC at point E, de ‖ BC intersects AB at point D, ∠ ade = 70 ° and calculates the degree of ∠ DEB Don't use the formula (one outer angle of a triangle is equal to the sum of two nonadjacent outer angles),
- 15. In the triangle ABC, ab = 15, AC = 20, the height ad on the side of BC = 12, try to find the length of BC? I give bonus points to the detailed process, get 7 within the 25 do not talk about I want the process of 7
- 16. There is a point a on the x-axis and a point B C (2,1) on the line y = X
- 17. BD bisects ∠ ABC, be bisects ∠ ABC into two parts of 3:4, ∠ DBE = 8 ° and calculates the degree of ∠ ABC
- 18. As shown in the figure, in △ ABC, AE is the angular bisector, and ∠ B = 52 °, C = 78 ° to find the degree of ∠ AEB
- 19. It is known that, as shown in the figure, in the triangle ABC, ab = AC, point D is the midpoint of BC, De is perpendicular to AB and E, DF is perpendicular to AC and F, so it is proved that angle DFE = angle def
- 20. In the triangle ABC, the angle ACB = 90 degrees, AC = BC, M is a point in the triangle ABC, and am = 3, BM = 1, CM = 2