![](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, △ ABC, ab = AC, D is on BC (D is not at the midpoint of BC), de ⊥ AB is on e, DF ⊥ AC is on F, BG ⊥ AC is on G, proving: de + DF = BG
It is proved that the area of △ ABC = the area of △ abd + the area of △ ACD, 12ab · de + 12ac · DF = 12ac · BG, ∵ AB = AC, ∵ de + DF = BG
As shown in the figure, in △ ABC, D is the point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
Prove: in △ ACD, because ad = AC and AE ⊥ CD, according to the intersection point of the vertical line and the bottom of the isosceles triangle, we can prove that e is the midpoint of CD, and because f is the midpoint of CB, so EF ∥ BD, and EF is the median line of △ BCD, so EF = 12bd, that is BD = 2ef
RELATED INFORMATIONS
- 1. As shown in the figure, in △ ABC, ab = AC, AE ⊥ AB in a, ∠ BAC = 120 ° AE = 3cm
- 2. In the triangle ABC, the angle ABC = 45 degrees, ad vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F, connecting cf. if the angle BAC is an acute angle In the triangle ABC, the angle ABC = 45 degrees, ad is vertical BC. The perpendicular foot is d. be vertical AC. the perpendicular foot is e. ad intersects be with F and connects cf. if the angle BAC is an acute angle, it is proved that the triangle CDF is an isosceles right triangle
- 3. Two forces F1 and F2 acting on the same object, on the same line, their resultant force is 20n. It is known that F2 = 50N, and the direction of F2 is the same as that of resultant force
- 4. Under the action of traction F1 and resistance F2 in the horizontal direction, when F1 = F2, F1 is less than F2 and F1 is greater than F2, the car will be ······························································································································?
- 5. In the test of verifying Newton's second law, why is the mass of the heavy object much less than that of the car?
- 6. In the test of verifying Newton's second law, why do we sometimes make the weight mass m far less than the trolley mass m? Sometimes we don't need to?
- 7. If the orbit radius of the earth around the sun is R1 and the revolution period is T1, and the orbit radius of the moon around the earth is R2 and the revolution period is T2, then the mass ratio of the sun to the earth is 0___ ．
- 8. If the mass ratio of the two planets is M1: M2 = P and the orbit radius ratio is R1: R2 = q, then the ratio of the two planets to the sun's gravity is F1: F2=______ ．
- 9. The radius of a planet is R1, the autobiographical period is T1, it has a satellite, the orbit radius is R2, the cycle around the planet is T2, if the gravitational constant is g, find the average density of the planet
- 10. 7. There are two forces F1 and F2 acting on the same object. When the directions of F1 and F2 are the same, the resultant force is 100N; when the directions of F1 and F2 are opposite, the resultant force is 100N The size is 50N, and the direction is the same as that of F1
- 11. As shown in the figure, it is known that ab ‖ CD, BD bisection ∠ ABC intersects AC in O, CE bisection ∠ DCG. If ∠ ace = 90 °, please judge the position relationship between BD and AC and explain the reason
- 12. As shown in the figure, in isosceles RT △ ABC, AC = BC, e is a point on the extension line of BC, connecting AE, if the vertical line of AE intersects the bisector of ACB at P, AC intersects F (2) Try to judge the quantitative relationship among BC, CE and CP (can you write it in the way of grade two) (3) If BC = 7, when CE = ----, AF = ef (fill in the conclusion directly)
- 13. As shown in the figure, in RT △ ABC, ∠ C = 90 °, a = 30 °, the bisector of ∠ C intersects with the bisector of the outer corner of ∠ B at point E, connecting AE, then ∠ AEB =?
- 14. In △ ABC, ∠ a = 40 °, the bisectors BD and CE of ∠ B and ∠ C intersect at m, then ∠ BMC
- 15. In the triangle ABC, a = 2 root sign 3, B is equal to 2 root sign 2, B = 45 ° to find a A / Sina = B / SINB = C / sinc 2×3½／sinA=2×2½／sinB sinA=3½／2 A = 60 ° or 120 ° I know it's 60 degrees. That's OK, but where does 120 degrees come from
- 16. If the area of △ CDE is 3, the area of △ BCE is 4, and the area of △ AED is 0 sorry... It's not complete.. AED is 6... It's Abe, but it's not trapezoid... And there were pictures... But I can't pass it. Please be complete. How can Z prove the similarity? But the two diagonals are not necessarily vertical...
- 17. As shown in the figure, in the quadrilateral ABCD, the bisector be of angle ABC intersects CD at e, the bisector CF of angle BCD intersects AB with F, be and CF intersect o, angle a = 124 degrees, angle d = 100 degrees, calculate the angle BOF 0 why no one
- 18. As shown in the figure, in △ ABC, ∠ C = 90 °, ad is the bisector of ∠ BAC, de ⊥ AB is on e, f is on AC, BD = DF. The following results are obtained: (1) CF = EB; (2) CBA + ∠ AFD = 180 °
- 19. On the opposite sides of the triangle ABC, ABC is known as a = 2 times root, 3C = 21 + Tana / tanb = 2C / B, and the area s of the triangle ABC is calculated
- 20. The edges corresponding to ABC are the edges corresponding to angle ABC, acosb + bcosa= The process is detailed