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In ⊙ o, the chord ab ⊥ CD is in M. the radius of ⊙ o is 5, BM = 6, am = 2. Find the size of dm-cm
O is op ⊥ AB is at P, O is OQ ⊥ DC is at Q,
According to the theorem, AP = BP = (6 + 2) / 2 = 4, DQ = (DM + cm) / 2
According to the Pythagorean theorem, Op = 3
OP = QM = 3 in rectangular OQMP,
QM=DM-DQ=DM-(DM+CM)/2=(DM-CM)/2
QM=3
Dm-cm = 6
In ⊙ o, string ab ⊥ string CD is in M, known ⊙ o radius is 5, BM = 6, am = 2, find the size of dm-cm (velocity ~ ~) I think some of the answers are as follows: O is op ⊥ AB is in p; O is OQ ⊥ DC is in Q, According to the theorem, AP = BP = (6 + 2) / 2 = 4, DQ = (DM + cm) / 2 According to the Pythagorean theorem, Op = 3 OP = QM = 3 in rectangular OQMP, QM=DM-DQ=DM-(DM+CM)/2=(DM-CM)/2 QM=3 Dm-cm = 6 But I think: BP = 4, BM = 6, Then MP = 2, so OQ = 2, In RT △ OQD, QD = Radix 21 Therefore, dm-cm = 2 times root number 21 - (root number 21-3) =Radical 21 + 3 == I looked at other people's answers and thought it was right, but I didn't think I could find any problems, Solution~~~~~~~~~
Analysis:
In your way, QD = √ 21, that's right
But there are some mistakes in DM and cm
DM = MQ + QD = OP + QD = 3 + √ 21 (when you think DM = 2mq, actually DC = 2mq > DM)
CM=CQ-MQ=DQ-OP=√21-3
So dm-cm = (3 + √ 21) - (√ 21-3) = 6
As shown in the figure, two chords AB and CD intersect vertically with m, am = 2, BM = 4, CM = 3, DM = 7. Find the radius of circle o
√26
In circle O, the chord AB intersects the chord CD in M, BM = DM, and am = cm
I'm not sure. Connect BD because BM = DM, so angle B = angle D, so arc BC = arc ad arc DC = AB, so chord AB = DC
So we're done
If M is a point in 0o, use a ruler to make a chord AB so that ab passes through point m and am = BM. How to draw a drawing? It is better to have a picture to explain
Connect OM, make the vertical line of OM through M, and intersect o with a and B, as required by your question
As shown in the figure, M is a point in ⊙ O. use ruler to draw a chord AB, so that ab passes through point m, and am = BM The process should be specific!
First of all, because a and B are on the circle, so Ao = Bo, am = BM, OM are common sides, so the two triangles are completely equal, so ∠ OMA = ∠ OMB = 90 ° so as to draw a chord with m as the perpendicular foot and perpendicular to OM, which is the required ab
As shown in the figure, △ ABC, ab = AC, O is a point in △ ABC, and ∠ OBC = ∠ OCB
It is proved that: ∵ AB = AC, ABC = ∠ ACB (equilateral and equal angle),
∵∠OBC=∠OCB，
Ψ ABO = ∠ ACO, OB = OC (equiangular to equilateral),
∴△AOB≌△AOC（SAS），
∴∠OAB=∠OAC，
And ∵ AB = AC,
⊥ BC
The chord CD of circle O has vertical diameter AB, and the perpendicular foot is p, Pa. the length of Pb is the square of x minus the two roots of 8 x + 12?
The intersecting string theorem PA * Pb = PD * PC and PD = PC, so PD = 2 root sign 3, so under ad = root sign (PA square + PD Square), there are two solutions ad = 4 or 4 root sign 3
As shown in the figure, in the circle O, AB is the diameter, CD is a chord, and CD ⊥ AB, the perpendicular foot is point P, connecting BC and ad. it is proved that the square of PC = PA * Pb
It is proved that because CD ⊥ AB, the perpendicular foot is point P and ab is the diameter
So PC = PD, and angle APC = angle BPD = 90 degree angle PAC = angle PDB angle PBC = angle pad
So PC / PA = Pb / PD
That is, PC * PD = PA * Pb
pc*pc=pa*pb
pc^2=pa*pb
As shown in the figure, in ⊙ O, AB is the diameter of ⊙ O, CD is a string, and CD ⊥ AB is connected to BC and AD at point P. to verify PC2=PA*PB
Connect AC and BC to get right triangle APC and BPC are similar triangles
PC / PA = Pb / PC
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