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As shown in the figure, a, B and C are three points that are not on the same straight line, AA ′‖ BB ′‖ CC ′, and AA ′ = BB ′ = CC ′. Prove: plane ABC ‖ plane a ′ B ′ C ′
It is proved that: ∵ AA ′ = BB ′, AA ′‖ BB ′, ∵ a ′ B ′‖ AB are parallelogram, ∵ a ′ B ′‖ AB, ab ⊂ face ABC ∩ a ′ B ′‖ face ABC, same as B ′ C ′‖ face ABC, ∩ a ′ B ′ C ′ = B ′, ∩ face ABC ‖ face a ′ B ′ C ′
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- 2. It is known that in the triangle ABC, O is any point in the triangle, and the extension lines of Ao, Bo, CO intersect the opposite sides of D, e, F Verification: AE / EC + AF / FB = AO / OD
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- 4. If O is the outer center of the acute triangle ABC and Ao, Bo and co intersect with L, m and N respectively, then Ao: Al + Bo: BM + CO: CN=
- 5. It is known that the triangle ABC is an isosceles triangle with an angle of 3 = 105 ° and the degree of angle 2 and angle 1 On the course evaluation
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- 8. If ∠ a = 4 ∠ B in the isosceles triangle ABC is known, the degree of ∠ C can be obtained
- 9. P is a point on the BC side of the triangle, and PC = Pb. Given that the angle ABC is equal to 45 degrees and the angle APC is equal to 60 degrees, calculate the degree of the angle ACB?
- 10. As shown in the figure, P is a point on the side BC of △ ABC, and PC = 2PB. It is known that ∠ ABC = 45 ° and ∠ APC = 60 °, then the degree of ∠ ACB is___ °．
- 11. There is a point O in the equilateral triangle ABC, the angle AOB = 113 ° and the angle BOC = 123 ° to find the degree of the three internal angles of the triangle with AO, Bo and CO as sides
- 12. In the triangle ABC, ab = AC, O is a point in the triangle ABC, angle OCB = angle OBC, and angle a = 40
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- 14. Given that OA = (- 1,8) ob = (- 4,1) OC = (1,3) in the rectangular coordinate plane, it is proved that △ ABC is an isosceles triangle
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