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Anonymous users2024-02-08Proof: (1) Nexus OC
ae is the diameter of o, and point c is the point above o.
oc=oa= ae
oac=∠oca
AC splits PAE equally
dac=∠oac
dac=∠oca
pa∥occd⊥pa
cd oc point c is o point up.
cd is the tangent of o.
2) Pass the point o as of pa in f
cd⊥pa，cd⊥oc
The quadrilateral CDFO is rectangular.
df=co=oa = ae，of=dc
ae=10df=co=oa=5
Let da=x, dc+da=6, da+af=dfdc=6-x, af=5-x
In RT AFO, af +of =oa, (5-x) +6-x) =5
x²-11x+18=0
x1=2，x2=9
x2=9 is not on topic, give up.
da=2af=5-x=3
The point o is the center of the circle, of ab
af= ab
ab=6
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Anonymous users2024-02-07Angular CAD is equal to angular CAE, so right-angle DAC is similar to right-angle CAE, i.e. angular CEA = angular DCA, i.e. CD is a circular O tangent.
dc+da=ad+of=6
af^2+of^2=25
af+ad=5
Solve the above three equations to obtain af
ab=6 can be obtained
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Anonymous users2024-02-06acd aec, so acd= aec= oce, dco= dca+ aco= oce+ aco+ aco=90
So oc cd, so cd is a tangent of .
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Anonymous users2024-02-05If the point c is used to make CF perpendicular AE, it is easy to know that the triangle CFA and the triangle CDA are congruent, that is, CF+AF=6, and we know AO=5
Let af=x, then there is cf 6 x and fo=5-x. Using the Pythagorean theorem in the right-angled triangle CFO: the square of (6-x) + the square of (5-x) = 5, after solving the equation, we get x=2 (x=9 rounded) that is, we get da 2, and if we connect co and extend it to intersect with be at the point g, then there is db cg, then there is equation 2 ab 5 ab 2 holds, and ab 6 is solved
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Anonymous users2024-02-04In the triangle AFO, af+1=fo, and from af 2+fo 2=25, af=3 and fo=4 are solved, so ab=2af=6
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Anonymous users2024-02-03Question 1: Both sides are multiplied by 2x(x-2), or the denominator is swapped to the other side of the equation.
In the second question, just multiply the 1-x 3-x on the left side by -1 to get the same denominator as on the right, and then multiply x-3 on the 2nd side.
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Anonymous users2024-02-021) y=(m-3)x+m -9, passing through the origin (0,0), indicating that when x=0, the plex row y=0, that is, 0=m -9, the solution is m=3 or m=-3, and because the coefficient of the first seepage noising term of a function cannot happen to be 0, m-3≠0, that is, m≠3, in summary, m=-3.
2) If y increases with the increase of x, then m-3 >0, and m>3 is obtained
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