Solution: (1) Point D is the golden section point on the side of AB, which is proved as follows:
∠A=36 0 ,AB=AC,∴∠B=∠ACB=72 0 .
∫CD average score ∠ACB, ∴∠DCB=36 0. ∴∠BDC=∠B=72 0 .
∵∠A=∠BCD,∠B=∠B,∴△BCD∽△BAC。 ∴ 。
bc = cd = ad,∴.
Point D is the golden section on the side of AB.
(2) The straight CD is the golden section of △ABC, which is proved as follows:
Let the height on the AB side of △ABC be h, then,
∴ 。
∫D is the golden section of AB. ∴ 。
∴ The straight CD is the golden section of △ABC.
(3)GH is not the golden line of right-angled trapezoidal ABCD, which is proved as follows:
∵BC∥AD,∴△EBG∽△EAH,△EGC∽△EHD。 ∴ 。
That's it.
Similarly, from △BGF∽△DHF, △CGF∽△AHF, that is.
∴ 。 ∴AH=HD。 ∴BG=GC。
∴ Trapezoidal ABGH and trapezoidal GHDH have equal bottoms and heights.
∴ 。
∴GH is not the golden section of right-angled trapezoidal ABCD.
(1) is proved by the relationship between the angles and sides of an isosceles triangle, and is obtained according to △BCD∽△BAC.
(2) According to the definition of golden section, it is proved that the straight CD is the golden section of △ABC.
(3) Repeatedly applying the similarity ratio of similar triangles, it is concluded that the upper and lower bottoms of trapezoidal ABGH and trapezoidal GHDH are equal respectively, and GH is not the golden section of right-angled trapezoidal ABCD.
(1) What are the main courses for junior high school students now?
Subjects Chinese, Mathematics and English.
Minor in physical ch