The angle formed by the intersection of a secant and a tangent is equal to the sum of the measures of the intercepted arcs.
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The arcs intersected by these chords are not congruent.
Given that two circles with different radii have chords AB and CD, such that AB is congruent to CD.
Let r₁ and r₂ be the radii of two different circles with centers O and O' respectively.
Assuming that the each of the ∠АОВ and ∠CO'D is less than or equal to π.
Then, we have isosceles triangle AOB and CO'D such that,
AO = OB = r₁,
CO' = O'D = r₂,
Let us assume that r₁< r₂;
We can see that arc(AB) intersected by AB is greater than arc(CD), intersected by the chord CD;
arc(AB) > arc(CD) .......(1)
Indeed,
arc(AB) = r₁ angle (AOB)
arc(CD) = r₂ angle (CO'D)
So, we have to prove that ;
∠AOB >∠CO'D ......(2)
Since each angle is less than or equal to π, and so
∠AOB/2 and ∠CO'D/2 is less than or equal to π
it suffices to show that :
tan(AOB/2) >tan(CO'D/2) ......(3)
From triangle AOB :
tan(AOB/2) = AB/(2*r₁)
tan(CO'D/2) = CD/(2*r₂)
Since AB = CD and r₁ < r₂ (As obtained from the result of (3) ), therefore, arc(AB) > arc(CD).
Hence, for two circles with different radii have chords AB and CD, such that AB is congruent to CD but the arcs intersected by these chords are not congruent.
Learn more about congruent from here brainly.com/question/1675117
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