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Point G is the intersection of the three middle lines of triangle ABC, Ag is perpendicular to GC, AC is equal to 4, and BG is obtained
BG=4
The vertical line of CG can be made through point B, and the perpendicular foot is point E
Prove that △ ACG ≌ △ beg
We know that the intersection point of the three middle lines of a triangle is also the center of gravity of the triangle. As shown in the figure, point G is the center of gravity of △ ABC, proving that Ag = 2Gd
It is proved that: as shown in the figure, crossing point D as DH ∥ AB and CE at h, ∵ ad is the middle line of △ ABC, ∥ point D is the middle point of BC, ∥ DH is the median line of △ BCE, ∥ be = 2dh, DH ∥ AB, ∥ CE is the median line of △ BCE, ∥ AE = be, ∥ AE = 2dh, ∥ DH ∥ AB, ∥ AEG ∥ DHG = aedh = 2, ∥ Ag = 2Gd
In the right triangle ABC, the angle c = 90 °, G is the center of gravity of the triangle ABC, and Ag is perpendicular to CG
Verification: when AB = 12, find the length of Ag
Prolongation of CG to AB in D and prolongation of Ag to BC in E
Connecting De, point G is the center of gravity, then D and E are the middle points
So: dg / GC = de / AC = 1 / 2
The angle ACB is 90 degrees, so CD = AB / 2 = 6, DG (1 / 3) CD = 2
AG=√(AD^2-DG^2)=√(36-4)=4√2.
In Δ ABC, Mn ∥ BC, DN ∥ cm, am & # 178; = ab · ad
D is on am, right?
∵MN∥BC,∴ΔAMN∽ΔABC,
∴AM/AB=AN/AC,
∵DN∥CM,∴ΔADN∽ΔAMC,
∴AD/AM=AN/AC,
∴AM/AB=AD/AM,
∴AM^2=AB*AD.
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