In the triangle ABC, D is the midpoint of AB, if AC = 15, BC = 8. CD = 8. 5, it is proved that the triangle ABC is a right triangle
Extend CD to F, make FD = 8.5, join dB, then BD = AC = 17 (parallelogram). Then 17 ^ 2 = 15 ^ 2 + 8 ^ 2. Then CFB is a right triangle. Because acbf is a parallelogram, ABC is a right triangle
As shown in the figure, in the triangle ABC, ad is perpendicular to point D, angle B = 45 degrees, angle c = 30 degrees, BC = 60, find the length of AD
30(√3-1)
RELATED INFORMATIONS
- 1. In △ ABC, if AB = 3, a = 45 ° and C = 75 °, then BC = () A. 3−3B. 2C. 2D. 3+3
- 2. Draw a plane rectangular coordinate system and calculate the area 1 a (- 6,0) B (0,8) C (0,3) s △ ABC 2 a (6,0) B (0,8) C (- 4,0) s △ ABC 3 A (- 4,3) B (- 4,2) C (5,2) for s △ ABC 4 a (5,5) B (- 3, - 3) C (16,0) where AB passes through the origin, find s △ ABC The area of sabcd can be calculated from the vertex a (0,0) B (2,5) C (7,7) d (4,0) of 5 ABCD quadrilateral
- 3. There are a (0,2), B (4,0), C (0,0), D (3,2) in the plane rectangular coordinate system. The equation for finding the circumscribed circle m of triangle ABC
- 4. There is a (0,1) B (2,1) C (3,4) d (2,4) 1 in the plane rectangular coordinate system There is a (0,1) B (2,1) C (3,4) d (2,4) 1 in the plane rectangular coordinate system. 2. If the midpoint of the chord cut by the circle m is exactly the point D, the equation of the straight line is obtained
- 5. As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° AB = 6, P is the point on AB, connect PC, set ∠ BCP = m ∠ ACP, when AP = 3 / 2, whether there is a positive integer m Make PC perpendicular to ab? If it exists, calculate the value of M. if it does not exist, explain the reason
- 6. P is a point in the equilateral triangle ABC, PC = 5, PA = 3, Pb = 4, find the degree of angle APB
- 7. In equilateral △ ABC, point P is in △ ABC, point q is outside △ ABC, and ∠ ABP = ∠ ACQ, BP = CQ. What shape of triangle is △ Apq? Try to state your conclusion
- 8. 1. As shown in the figure, a is the point on ⊙ o, the extension line of tangent intersection diameter CB passing through point a is at point P, ad ⊥ BC is at point D 1. As shown in the figure, a is the point on ⊙ o, the extension line of tangent diameter CB passing through point a is at point P, ad ⊥ BC is at point D Verification: Pb / PD = PO / PC Indicate BDO three points collinear
- 9. As shown in the figure, it is known that AB is the diameter of ⊙ o, Pb is the tangent of ⊙ o, B is the tangent point, Op ⊥ chord BC is at point D and intersects ⊙ o at point E. (1) prove: ∠ OPB = ∠ AEC; (2) if point C is the triad point of semicircle and ACB, please judge which special quadrilateral AOEC is? And explain the reason
- 10. The isosceles triangle ABC, with a point D on the waist AB, connects DC and makes the isosceles triangle EDC with DC as the bottom edge, which is similar to the triangle ABC, connects AE and proves that AE is parallel to BC
- 11. The isosceles triangle ABC AB = AC, the angle BAC = 45 degrees, ad vertical BC to D, CE vertical AB to e, ad and EC intersect h, prove that ah is equal to BC
- 12. Let a, B and C be the lengths of three sides of the triangle ABC, and prove that x + 2aX + B = 0 and X + 2cx - B = 0 have common roots if and only if the angle a = 90 degree
- 13. If two right sides a and B of a right triangle satisfy a + B = 7 and hypotenuse C = 6, then the area of the right triangle is?
- 14. A right triangle, three variable length are 6cm, 8cm, 10cm, the area of this triangle is (), the height of the hypotenuse is
- 15. The length of the hypotenuse of a right triangle is 10 cm, and the sum of the two right sides is 12 cm? Pythagorean theorem should be used
- 16. In the triangle ABC, a = 18, B = 16, a = 150, solve the triangle
- 17. (1) In triangle ABC, a = 2, B = 2 √ 3, B = 60 ° are known to solve triangle ABC (2) In triangle ABC, we know a = √ 3, B = √ 2, B = 45 ° to solve triangle ABC
- 18. In the triangle ABC, a = 3, B = root 3, a = 60 degree solution triangle
- 19. As shown in the figure, the bisectors of ∠ CBD and ∠ BCE at the two outer angles of △ ABC intersect at point O, ∠ a = 40 ° to calculate the degree of ∠ BOC
- 20. As shown in the figure, OE bisects ∠ BOC, OD bisects ∠ AOC, ∠ BOE = 20 ° and ∠ AOD = 40 ° to calculate the degree of ∠ doe