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A geometry problem in the third grade of junior high school In the circle O, the diameter AB intersects with the diameter CD, passing through three points a, O and B respectively to make the vertical line of CD, and the vertical feet are e, h and f respectively
It is proved that a, O and B are perpendicular to CD, and the perpendicular feet are e, h and f respectively,
So AE parallel Oh parallel BF,
So eh = FH,
According to the vertical diameter theorem,
CH=DH,
So ch-eh = dh-df
So CE = DF
As shown in the figure, e, G, F and H are the points on the four sides of rectangle ABCD, EF ⊥ GH. If AB = 2 and BC = 3, then EF: GH = ()
A. 2:3b. 3:2c. 4:9d. Not sure
If f is used as FM ⊥ AB in M and h as HN ⊥ BC in N, then ⊥ 4 = ⊥ 5 = 90 ° = ⊥ AMF ∵ quadrilateral ABCD is a rectangle, ⊥ ad ∥ BC, ab ∥ CD, ⊥ a = ⊥ d = 90 ° = ⊥ AMF, ⊥ quadrilateral amfd is a rectangle, ⊥ FM ∥ ad, FM = ad = BC = 3, similarly, HN = AB = 2, HN ∥ AB, ∥ 1 = ⊥ 2, ≁ Hg ⊥ EF, ≁ ho
A mathematical geometry problem in the third grade of junior high school
In △ ABC, AC = 2dc, ad = √ 5dc
(1) Find the degree of ∠ C
(2) lengthen CA to F, make AF = CD, lengthen CD to g, make DG = CF, Ag and FD intersect at point h, and verify: ∠ GHD = 45 °
(3) On the basis of (2), if CD = 1, please write the value of ah as_______ .
Let DC be x, so AC = 2x ad = √ 5x, so the square of AC + the square of CD = the square of AD, so the angle c = 90 degrees
I don't understand
As shown in the figure, in △ ABC, ad is the middle line on the side of BC, just a triangle, and then ad is the middle line of BC,
(1) Explore the relationship between AB, AC and ad, and explain the reason
The answer is shown in the picture!
To solve the following geometric problems (third grade), urgent!
If BC = 2, AC = 3, then AE · EB=————
Let the radius of the inner loading circle be r
Then AE = 3-R, EB = 2-r
AE*EB=6-5R+R^2
And AE + EB = √ (9 + 4) = √ 13
(3-R)+(2-R)=√13
R=(5-√13)/2
AE*EB=6-5R+R^2
=6-5(5-√13)/2+((5-√13)/2)^2
=3
Geometry proof in junior three!
As shown in the figure, in the isosceles triangle ABC, CA = CB, ∩ ACB = 90 °, points D and E are two points on the straight line BC and CD = be. Passing through point AC, make cm ⊥ AE, intersect AE at point m, intersect AB at point F, connect DF and extend AE at point n
(2) And the verification: ∩ d = ∩ E
(2) It is proved that the angle AGC is 90 degrees because CG is perpendicular to G and extending CG intersects AB at h. The triangle ABC is isosceles right triangle because angle ACB is 90 degrees and Ca is CB, so CG is the vertical line of isosceles right triangle ABC. The angle bisector is angle AGC = 90 degrees, angle ACh = angle BCH = 1 / 2, angle ACB = 45 degrees, so angle ACh = angle CBA =
It is known that, as shown in the figure, P is the interior point of the square ABCD, ∠ pad = ∠ PDA = 15 °. It is proved that △ PBC is an equilateral triangle
It is proved that: ∵ square ABCD, ∵ AB = CD, ∵ bad = ∵ CDA = 90 °, ∵ pad = ∵ PDA = 15 °, ∵ PA = PD, ∵ PAB = ∵ PDC = 75 °, congruence of △ DGC and △ ADP is made in the square, ∵ DP = DG, ∵ ADP = ∵ GDC = ∵ DAP = ∵ DCG = 15 °, ∵ PDG = 90 ° - 15 ° - 15 ° = 60 °, ∵ PDG is an equilateral triangle (an isosceles triangle with an angle equal to 60 degrees is an equilateral triangle),
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