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In trapezoidal ABCD, ab ∥ DC, diagonal AC, BD intersect with O, through o as Mn ∥ DC intersects ad with m, intersects BC with N. to prove the value of OM / AB + on / BC AB is the bottom, DC is the bottom, a and D are on the left
∵MN∥DC,AB∥DC
∴MN∥AB
∴∠DAB=∠DMO,∠DBA=∠DOM
∴ΔABD∽ΔMOD
OM/AB=MD/AD
Similarly, on / BC = BN / BC
From ab ∥ Mn ∥ CD, BN / BC = am / ad can be obtained
∴OM/AB+ON/BC=MD/AD+AM/AD=AD/AD=1
In the parallelogram ABCD, the diagonal lines AC and BD intersect at the point O, and make any straight line Mn through the point O, then om = on? Why?
Second semester mathematics synchronous detection level 3 jump on the question!
Because the diagonals of parallelograms are equally divided, Ao = Co, the opposite vertex angles are equal, and a set of internal stagger angles are equal, △ AOM ≌ △ con (ASA), so om = on
In the cube abcd-a'b'c'd ', m and N are the middle points of edges AA' and ab respectively, and P is the center of the upper and lower surface ABCD, then the angle between the line Pb and Mn is ()
A. 30°B. 45°C. 60°D. 90°
First, draw a graph to translate Mn to A1B, ∠ a1bp is the angle formed by the straight line Pb and Mn. Let the side length of the cube be a, a1p = 22a, A1B = 2A, BP = 64a, cos ∠ a1bp = 32, and ∠ a1bp = 30 °, so select a
In the cube abcd-a'b'c'd ', M is the midpoint of AA'
Let the edge length of cube be 2 and an = X
Then MC '= 3
Mn = radical (x ^ 2 + 1)
C'n = root (12-4x + x ^ 2)
MN⊥MC'
Mc'^2+MN^2=NC'^2
9+x^2+1=12-4x+x^2
4x=2
x=1/2
Mn ⊥ MC '
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