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As shown in the figure, in ▱ ABCD, e and F are two points on the diagonal BD, and be = DF
It is proved that: ∵ quadrilateral ABCD is a parallelogram, ≌ AB = CD, ≌ Abe = ≌ CDF (SAS), ≌ BAE = ≌ DCF in △ Abe and △ DCF
As shown in the figure, in square ABCD, ab = 3, points E and F are on BC and CD respectively, and ∠ BAE = 30 ° and ∠ DAF = 15 ° to calculate the area of △ AEF
Rotate △ ADF 90 ° clockwise around point a to the position of △ ABG, ≌ Ag = AF, ≌ gab = ≌ fad = 15 °, ≌ gae = 15 ° + 30 ° = 45 °, ≌ EAF = 90 ° - (30 ° + 15 °) = 45 °, ≌ AE = AE, ≌ AEF ≌ AEG, ≌ EF = eg, ≌ AEF = ≌ AEG = 60 ° in RT △ ab
In the parallelogram ABCD, through point a, make AE vertical BC, AF vertical CD respectively, and prove the angle BAE = angle DAF
∠abe=∠adf
∠aeb=∠afd=90°
So BAE = DAF
In the triangle ABC, angle B = 2, angle c, ad perpendicular to BC and D, if extend CB to e, make be = AB, link AE, prove: CD = AB + BD
ABC=AEB+EAB AB=BE ABC=2AEB=2C AE=AC CD=DE =BE+BD=AB+BD
As shown in the figure, in △ ABC, D is the midpoint of the BC edge, e is any point on the AC edge, be intersects ad at point O, and AEEC = 1n
Let d be DF ∥ be, Ao: ad = AE: AF. ∵ d be the midpoint of BC edge, ∵ CF = EF = 0.5ec. ∵ AEEC = 1n, ∵ aeac = 1n + 1, namely AE: (AE + 2ef) = 1: (1 + n), ∵ AE + 2ef = AE + AEN, ∵ AE: EF = 2: n. ∵ aeef = aood, ∵ Ao: od = 2: n
As shown in the figure, △ ABC, points D and E are the midpoint of AB and AC respectively, connecting De, and segments be and CD intersect at point O. if od = 2, then OC=______ ．
Solution 1: ∵ points D and E are the midpoint of AB and AC respectively, line segments be and CD intersect at point O, ∵ o point is the center of gravity of △ ABC, ∵ OC = 2od = 4; solution 2: ∵ points D and E are the midpoint of AB and AC respectively, ∵ De is the median line of △ ABC, ∥ de ∥ BC, de = 12bc, ∵ ode = ∥ OCB, ∥ OECD = ∥ OBC, ∥ ode ∥ OCB, ∥ od: OC = de: BC = 1:2, ∥ OC = 2od = 4
If the triangle ABC is inscribed in circle O, D is the midpoint of line AB, extend OD, intersect circle at point E, and connect AE and be, then the following is true:
How many are 1 AB perpendicular to de 2 AE equal to be 3 od equal to de 4 angle AEO equal to angle c?
There are two correct answers, each of which is (1), (2) connecting OA ob, OA = ob, because D is the midpoint, so ad = BD, because od = OD, so triangle AOD is equal to triangle BOD, so angle ADO = angle BOD = 90 degrees, so De is the vertical line of AB, so AE = be
As shown in the figure, the equilateral triangle ABC, AE, CD are perpendicular to the plane ABC, AE = AB = 2CD = 2A, f is the midpoint of be, DF is parallel to the plane ABC
Second question: find AF perpendicular to BD
emergency
Second, let's see: ([] stands for the root)
Connect BD, ad and AF, take the midpoint m, Q and P of AD, AC and BF, and connect NP, PM and Mg
It can be seen from the above that PN / / = 1 / 2AF, Mn / / = 1 / 2bd
Because AE = AB = 2CD = 2A, it can be calculated that
NP=[2]/2,MP=[5]/2,PM=[7]/2;
In accordance with Pythagorean theorem, that is proved
=As shown in the figure, it is known that the three vertices of △ ABC are on the circle with o as the center, ad is the height of △ ABC, AE is the diameter of the circle with o as the center, and ab × AC = ad × AE is proved
Connect be
∵ AE is the diameter of circle o
∴∠ABE=90°
∵ ad is the high value of △ ABC
∴∠ADC=90°
In △ Abe and △ ADC,
The angle of circle opposite to the arc is equal
∴△ABE∽△ADC
∴AB/AD=AE/AC
∴ABXAC=ADXAE
Figure you should have it, I draw according to their own diagram, there are different tell me
Ad is the height of △ ABC, A.B.C is on circle O, AE is the diameter of circle o
Please write the process
This problem is to prove that ab: AE = ad: AC
First, make an auxiliary line, that is, connect CE. Now there are two right triangle lines △ abd and △ AEC. Angle B and angle e are equal. Because they are on the same screw, right angles △ abd and △ AEC are similar
Then the corresponding edges of similar triangle lines are proportional
Learn to refuel:)
RELATED INFORMATIONS
- 1. The quadrilateral ABCD and the quadrilateral aefg are both rhombic. The point E is on the ad side and the point G is on the extension line of the Ba side As shown in Figure 1, BD and CE are their diagonals. The extension line of Ge intersects BD at point Q and DC at point M. verify: GM ⊥ BD
- 2. The square ABCD is rotated clockwise around point a to obtain the quantitative relationship of square aefg, the intersection of edge FG and BC, aefc, the intersection of edge FG and BC, h, Hg and Hb
- 3. The side length of the bottom of the hexagonal prism is equal to 1, and the side length is 2. After the side of the prism is expanded along a side edge, the area of the side expansion is calculated
- 4. If the front view and top view of a triangular prism with equal edge lengths are square and regular triangle respectively, then the left view is () A. Rectangle B. square C. diamond D. regular triangle
- 5. Red pigment + yellow pigment red pigment + blue pigment yellow pigment + blue pigment
- 6. English translation English
- 7. The side length of square ABCD is 4, M is the middle point of AD, the moving point e moves on the line AB, connects EM and extends the intersection line CD and F, and makes the intersection line of EF through M BC and G, connect eg and FG, prove that let AE = x, the area of △ EGF = y, find the analytic formula and value range of Y and X. can △ GEF be an equilateral triangle during the movement of 2 points e? Please explain the reason
- 8. The area of quadrilateral ABCD is 30, e and F are the three equal parts of AB, m and N are the three equal parts of DC. The area of efnm is calculated
- 9. The right figure ABCD is a right angled trapezoid. Take CD as the axis and rotate the trapezoid around this axis to get a rotating body. How many cubic centimeters is its volume?
- 10. 17. As shown in the figure, in the pyramid p-abcd, the bottom surface is a right angled trapezoid, ad ‖ BC, ∠ bad = 90 °, PA ⊥ bottom surface ABCD, and PA = ad = AB = 2BC, m, n respectively If the first question is proved, ask the second question. Please don't use vectors
- 11. A piece of rectangular paper with a length of 5cm and a width of 4cm is rotated around its length and width respectively to form two cylinders. Which cylinder has the largest volume should be estimated first and then calculated
- 12. The length of a rectangle is 5cm and the width is 4cm. After rotating around the short side, you will get a cylinder whose volume is () Chasing 50 points
- 13. The length of a rectangle is 10cm, the width is 3cm, take the length as the axis, rotate a circle to get a cylinder, calculate the volume of the cylinder
- 14. The length of a rectangle is 5cm and the width is 12cm. Taking one of the edges as the axis, the rectangle is rotated for one circle, and the volume of the three-dimensional figure is ()
- 15. The length and width of a rectangle are 8cm and 5cm respectively. The length and width of the rectangle are used for sleeve rotation (as shown in the figure) to form two cylinders The length of a rectangle is 8cm and the width is 5cm. The length and width of the rectangle are used for sleeve rotation (as shown in the figure) to form two cylinders. Compare the volume of these two cylinders, which one is larger? How much bigger?
- 16. E. F is the edge BC of square ABCD with side length 4, the point on CD, CE = 1, CF = 4 / 3, the extension line of line Fe intersecting AB is at g, passing through a moving point h on line FG, making HM perpendicular to Ag and m, let HM be x, and the area of rectangle amhn be y 1. Find the function expression 2 when x =? Amhn, what is the largest area When x = 3, the maximum value of Y is 12 How to - -,
- 17. As shown in the figure, in the trapezoid ABCD, ad is parallel to BC, and the bisector ce of angle BCD intersects the midpoint e of ab
- 18. As shown in the figure, △ ABC, e is the point on AC, and AE = AB, ∠ EBC = 12 ∠ BAC, the ⊙ o with ab as the diameter intersects AC at point D and EB at point F. (1) prove that BC is tangent to ⊙ o; (2) if AB = 8, sin ∠ EBC = 14, find the length of AC
- 19. Urgent! As shown in the figure, in △ abd and △ ace, F and G are the intersection points of AC and DB, AB and EC respectively. There are four propositions: ① AB = AC; ② ad = AE; ③ AF= As shown in the figure, in △ abd and △ ace, F and G are the intersection points of AC and DB, AB and EC respectively. There are four propositions as follows: ① AB = AC. ② ad = AE, ③ AF = AG, ④ ad ⊥ BD, AE ⊥ CE. Take three propositions as the topic and the other as the conclusion to construct a proposition. Write as much as you can. Add proof. Hurry. 5 minutes
- 20. As shown in the figure, the quadrilateral ABCD is inscribed in ⊙ o, BD is the diameter of ⊙ o, AE ⊥ CD, perpendicular foot is e, Da bisection ⊙ BDE