Please answer the 14th question
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As the chords are equidistant from the centre, they are equal.
Now construct the perpendicular bisector of the chord from the center. As AB=CB, the bisectors of the chords will be equal. Also the bisected lines will be equal and AB=AB (common) and therefore the triangles will be congruent. Therefore by CPCT we can prove that the diameter DB bisects the angle ABC.
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Now construct the perpendicular bisector of the chord from the center. As AB=CB, the bisectors of the chords will be equal. Also the bisected lines will be equal and AB=AB (common) and therefore the triangles will be congruent. Therefore by CPCT we can prove that the diameter DB bisects the angle ABC.
Thumbs up if you can understand it easily!!
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In figure,AB and CB are chords equidistant from centre O.
It means,AB and BC are of equal length
AB = BC
In triangles,ADB and CDB
BD=BD. [Common side]
AB=BC [Chords of equal length]
CD=AD
triangle ADB ≈ triangle CDB. [SSS congruency rule]
NOTE: ≈ —————→congruent to
angle ABC=angle ADC. [CPCT]
diameter DB bisects angle ABC and angle ADC.
hence proved.
please mark as brainliest answer
It means,AB and BC are of equal length
AB = BC
In triangles,ADB and CDB
BD=BD. [Common side]
AB=BC [Chords of equal length]
CD=AD
triangle ADB ≈ triangle CDB. [SSS congruency rule]
NOTE: ≈ —————→congruent to
angle ABC=angle ADC. [CPCT]
diameter DB bisects angle ABC and angle ADC.
hence proved.
please mark as brainliest answer
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