Question

prove that if chords of congruent circles substend equal angles at their centre then the chords are equal
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Answer 1

Answer:

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• Two circles are congruent
• Angle subtended at the center by the two chords are equal
• The chords are equal?

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

• Here let's take two circles of equal radius and two chords AB & A'B'
• Here we shall prove that the two triangles are congruent hence proving the given statement
• To prove that in general we shall use those two radius and one angle of the two triangles

In ∆AOB & ∆A'O'B'

$\displaystyle\sf :\implies OA = O'A' \:\: \bigg\lgroup Radius \ of \ congruent \ Circles\bigg\rgroup$

$\displaystyle\sf :\implies \angle AOB = \angle A'O'B' \:\: \bigg\lgroup Given\bigg\rgroup$

$\displaystyle\sf :\implies OB = O'B' \:\: \bigg\lgroup Radius \ of \ congruent \ Circles\bigg\rgroup$

$\displaystyle\sf \therefore \triangle AOB \cong \triangle A'O'B' \bigg\lgroup SAS \bigg\rgroup$

$\displaystyle\underline{\bigstar\:\textsf{From CPCT :}}$

$\displaystyle\sf \dashrightarrow AB = A'B'$

$\displaystyle\underline{\textsf{ \textbf{ Hence Proved!! }}}$