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As shown in the figure, ⊙ o is the circumscribed circle of △ ABC, the chord CD bisects ∠ ACB, ∠ ACB = 90 ° to verify: Ca + CB = 2CD
It is proved that: connecting ad, BD, passing a as am ⊥ CD, passing B as BN ⊥ CD, the perpendicular feet are m, n respectively, ∵ AB as diameter, CD bisects ⊙ ACB intersection ⊙ o to D, ∵ ACD = ∵ BCD = 12 ⊙ ACB = 45 °, both ∵ ACM and ∵ BCN are isosceles right triangles, ad = BD, in RT △ ACM, CM = 22ca, in RT △ BCN, CN = 22cb, ∵ cm + CN = 22 (Ca + CB), ∵ AB is diameter, ∵ ADB = 90 °, and ∵ ADM + BDN = 90 In △ ADM and △ BDN, ∠ ADM = ∠ DBN, in △ ADM and △ BDN, ∠ ADM = ∠ DBN & nbsp;; amd = ∠ DNB = 90 ° & nbsp; ad = BD & nbsp;, ≌ ADM ≌ BDN (AAS), ≌ DN = am, and ≌ am = cm (the right sides of isosceles right triangle are equal), ≌ cm = DN, ≌ CD = CN + DN = CN + cm = 22 (Ca + CB), ≌ Ca + CB = 2CD
In the triangle ABC, the angle ACB = 90 °, CB = CA = root 2, and the angle ECF = 45 ° are proved; BF multiplied by AE = 2
Because the angle ACB = 90 ° CB = ca
So angle B = angle cab = 45
Because angle EFC = angle efc
Angle B = angle ECF = 45 degree
So triangular BFC is similar to CFE
The same is true
Triangle ace is similar to CFE
So triangle ace is similar to triangle BFC
BF/AC=BC/AE
BF * AF = square of root 2 = 2
If the radius of circle O is r = 2, SINB = 3 / 4, then the length of chord AC is
How to think about this question?
Chord AC = 3
As shown in the picture
In the circle
Circumference angle = (1 / 2) circumference angle
so
∠AOC=2∠ABC=2∠AOD
sinB=3/4
therefore
sin∠AOD=3/4
AC=2AD=2×(AO×sin∠AOD)
AC=2×(2×3/4)
AC=3
There may also be another situation, such as
The obtuse angle is ABC, as shown in the figure where B 'is
But at this time, the chord of ∠ B is the same AC, so the result is the same
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