![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in the figure, in the rectangle ABCD, ab = 2, BC = 3, the vertical bisector of diagonal AC intersects ad respectively, BC connects CE at points E and F, then the length of CE is______ .
Therefore, AE = EC, Ao = Co
So △ AOE ≌ △ COE
Let CE be X
Then de = ad-x, CD = AB = 2
According to Pythagorean theorem, X2 = (3-x) 2 + 22 can be obtained
CE = 13
6．
So the answer is 13
6．
As shown in the figure, in the rectangle ABCD, ab = 2, BC = 4, the vertical bisectors of diagonal AC intersect ad AC at points E and O respectively, connecting CE and CE___
Suppose OE intersects BC with point F
It is known that AC = 2 √ 5. OC = √ 5
Because ∠ ABC = 90 ° = ∠ COF
So △ ABC is similar to △ FOC
So AB / BC = of / OC
It can be found that of = √ 5 / 2 = OE
So the length of CE can be calculated according to the length of OE and OC. The answer is c.2.5
As shown in the figure, in the rectangle ABCD, ab = 2, BC = 3, the vertical bisector of diagonal AC intersects ad respectively, BC connects CE at points E and F, then the length of CE is______ .
Therefore, AE = EC, Ao = Co
So △ AOE ≌ △ COE
Let CE be X
Then de = ad-x, CD = AB = 2
According to Pythagorean theorem, X2 = (3-x) 2 + 22 can be obtained
CE = 13
6．
So the answer is 13
6．
As shown in the figure, in the rectangle ABCD, ab = 2, BC = 3, the vertical bisector of diagonal AC intersects ad respectively, BC connects CE at points E and F, then the length of CE is______ .
Therefore, AE = EC, Ao = Co
So △ AOE ≌ △ COE
Let CE be X
Then de = ad-x, CD = AB = 2
According to Pythagorean theorem, X2 = (3-x) 2 + 22 can be obtained
CE = 13
6．
So the answer is 13
6．
As shown in the figure, in square ABCD, e is the midpoint of AD, BD and CE intersect at point F as shown in the figure. In square ABCD, e is the midpoint of AD, BD and CE intersect at point F. verification: AF ⊥ B E
Let AF intersect be at M,
DA=DC, ∠ADF=∠CDF=45°, FD=FD ==> △DAF≌△DCF ==> ∠DAF=∠DCF
AE=ED, ∠BAE=∠CDE=90°, AB=DC ==> △ABE≌△DCE ==> ∠BEA=∠CED
So ∠ DAF + ∠ bea = ∠ DCF + ∠ CED = 180 ° - ∠ CDE = 90 °
That is, ∠ EAM + ∠ mea = 90 °, so ∠ EMA = 180 ° - 90 ° = 90 °,
AF ⊥ be
I admire shuxpp, but he has a question in the last step, so he added
As shown in the figure, in the square ABCD, D is de ‖ AC, ∠ ace = 30 °, CA = CE, CE intersects ad at point F, proving that AE = AF
It is proved that: ∵ CA = CE, ∵ ace = 30
∴∠AEF=1
2（180°-∠ACE）=75°
∵ the quadrilateral ABCD is a square
∴∠CAD=45°
∴∠AFE=∠CAD+∠ACE=75°
∴∠AEF=∠AFE
∴AE=AF．
As shown in the figure, in the square ABCD, D is de ‖ AC, ∠ ace = 30 °, CA = CE, CE intersects ad at point F, proving that AE = AF
0
0
(1) Proof: connect AC as shown in the figure,
∵ point a is the midpoint of arc BC,
∴∠ABC=∠ACB，
And ∵ ACB = ∵ ADB,
∴∠ABC=∠ADB．
And ∵ BAE = ∵ BAE,
∴△ABE∽△ABD；
（2）∵AE=2，ED=4，
∴AD=AE+ED=2+4=6，
∵ Abe ∵ abd, BD is the diameter of ⊙ o,
∴∠BAD=90°，
∵△ABE∽△ABD，
∴AE
AB=AB
AD，
∴AB2=AE•AD=2×6=12，
∴AB=2
3，
In RT △ ADB, Tan ∠ ADB = 2
Three
6＝
Three
3．
As shown in the figure, AB is the diameter of circle O, e is a point on circle O, C is the midpoint of arc EB, CD is perpendicular to D. try to judge the position relationship between OC and AD I don't understand the answer on the Internet. Thank you for the detailed process
∵ AB is the diameter,
﹤ AEB = 90 ° (the circumference angle of the diameter is equal to 90 °)
Be ⊥ AE,
∵ C is the midpoint of the arc EB,
⊥ be (inverse theorem of the vertical diameter theorem)
/ / OC ∥ ad (two lines perpendicular to the same line are parallel)
The condition "CD perpendicular to D" is redundant
As shown in the figure, AB is the diameter of the circle O, C is the midpoint of the semicircle, and D is a point on the arc AC. extend ad to e so that AE = BD, connect CE, and calculate CE / De
First of all, we should make three auxiliary lines, which connect CD, CB and AC respectively. Then we can know from the meaning of the title that ∠ ACB is 90 ° and C is the midpoint of arc AB, so AC = BC and can be obtained from the same arc's circumferential angles, ∠ EAC = ∠ CBD, and from the meaning of the title, AE = BD can be obtained by the definition of edges, corners and edges, △ EAC ≌ △ DBC, so CE = CD, △ CED is isosceles
RELATED INFORMATIONS
- 1. As shown in the figure, BD is the diameter of ⊙ o, point a is the midpoint of arc BC, ad intersects BC at point E, AE = 2, ed = 4 (1) The results were as follows: △ Abe ∽ abd; (2) Find the value of Tan ∠ ADB; (3) Extend BC to f and connect FD so that the area of △ BDF is equal to 8 3. Find the degree of ∠ EDF
- 2. As shown in the figure, in ⊙ o, D and E are points on the radius OA and ob respectively, and ad = be. Point C is a point on arc AB, connecting CD, CE, Co, ∠ AOC = ∠ BOC Confirmation: CD = CE
- 3. As shown in the figure, given that the radius of ⊙ o is 1, AB and ⊙ o are tangent to point a, OB and ⊙ o intersect at point C, CD ⊥ OA, and the perpendicular foot is D, then the value of cos ⊥ AOB is equal to () A. OD B. OA C. CD D. AB
- 4. It is known that AB and CD are the two diameters of ⊙ o, and the chord CE ∥ ab. it is proved that ad = AE
- 5. As shown in the figure, circle O and circle 1 intersect at points a and B, AC is the diameter of circle O, and the extension lines of Ca and CB intersect circle O1 at points D and E, AC = 12, be = 30, BC = ad, Find ∠ C and de
- 6. AB is the diameter of ⊙ o, CD, CB are the tangent lines of ⊙ o, B and D are tangent points, connecting OC to ⊙ o to e, straight line AE to BC, BD to F, G. verification: AF bisection ⊙ bad
- 7. As shown in the figure, a is a point outside the circle O with radius 1, OA = 2, AB is the tangent line of ⊙ o, B is the tangent point, the chord BC ∥ OA, connecting AC, then the area of the shadow part is equal to () A. Three Four B. π Six C. π 6+ Three Eight D. π 4− Three Eight
- 8. As shown in the figure, given that D is a point on the extension line of edge BC of △ ABC, DF ⊥ AB is at point F, intersects AC at point E, ∠ a = 40 ° and ∠ d = 30 °, then the degree of ∠ ACB is obtained______ Degree
- 9. As shown in the figure, ad is the angular bisector of △ ABC, de ⊥ AB, DF ⊥ AC, and the perpendicular feet are e and f respectively. If AB = 5, AC = 3, and the area of △ ABC is 16, find the length of de?
- 10. As shown in the figure, △ ABC, ab = AC, e is a point on AB, f is a point on AC extension line, and be = CF, if EF and BC intersect at D, prove: de = DF
- 11. It is known that in △ ABC, D and E are points on BC and ab respectively. AD and CE intersect F and CD = 1 3BC，AE=2 5ab. Find s △ ACF The value of s △ CDF
- 12. As shown in Figure AB is the diameter of circle O, ad and circle O are tangent to point a, de and circle O are tangent to point E, and point C is a point on the extension line of De, and CE = CB is connected to the extension of AE and AE
- 13. As shown in the figure, ad vertical BC BD = DC point C on the vertical bisector of AE, what is the relationship between the length of AB AC CE and ab + BD and de 1. As shown in the figure, ad vertical BC, BD = DC, point C is on the vertical bisector of AE, what is the relationship between the length of AB AC CE and ab + BD and de?
- 14. As shown in the figure BD is the diameter of circle O, ab = AC, ad intersects BC at point E, AE = 2, ed = 4 The first question is to find the length of AB, the second question is to extend DB to f so that BF = Bo, connect FA, and prove that FA is tangent to circle o
- 15. Known, as shown in the figure, ab ‖ CD, AE bisection ∠ BAC, CE bisection ∠ ACD, find the degree of ∠ E
- 16. If the tangent point of a circle is a tangent point of a circle, then the tangent point of the circle is abo Take yourself in the picture below
- 17. In the triangle ABC, ab = ac.f on AC, AE = AF is intercepted on the extension line of Ba to verify EF vertical BC
- 18. As shown in the figure, AB is the diameter of ⊙ o, C is ⊙ o, and the tangent line passing through point C is perpendicular to each other, and the perpendicular foot is D, so as to verify the AC bisection angle DAB As shown in the figure, AB is the diameter of ⊙ o, C is the point on ⊙ o, CD cuts ⊙ o at point C, [label: ab CD, diameter, CD]
- 19. Given that the circle C (x + radical 3) ^ 2 + y ^ 2 = 16, point a (radical 3,0) q is a moving point on the circle, the vertical bisector of AQ intersects CQ at point m, and the trajectory equation of M is e Find the equation of E. urgent! The area of the l-intersection locus e of the straight line passing through the point P (1,0) at two different points a, B and △ AOB is 4 / 5
- 20. P is a point outside the circle O, Pb is the tangent line of circle O, a and B are the tangent points, BC is the diameter of circle O, so AC is parallel to Po As the title