# Answer to Question #4126 in Geometry for john allen

Question #4126

Two circles with different radii have chords AB and CD, such that AB is congruent to CD. Are the arcs intersected by these chords also congruent? Explain.Consider the type of triangle that may be drawn by connecting the endpoints of a chord to the center of a circle.Compare the triangles made by two circles with different radii.

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Expert's answer

**Solution.**

Let R

_{1}and R

_{2}be the corresponding radii of circles, and O, T be the centers of the circles.

Notice that it is implicitly assumed that assuming that each of the AOB and CTD is less of equal to pi.

Then we have a isosceles triangles AOB and CTD such that

OA=OB=R

_{1},

TC=TD=R

_{2}.

For definiteness assume that

R

_{1}<R

_{2}.

We claim that the arc A^B intersected by AB is greater than C^D, the arc intersected

by CD:

(*) A^B > C^D

Indeed,

A^B = R1 angle(AOB),

C^D = R2 angle(CTD).

So instead of (*) we have to prove that

(**) angle(AOB) > angle(CTD).

Moreover, since each of these angles is less of equal to pi, and so

angle(AOB)/2, angle(CTD)/2 is less of equal to pi/2,

it suffices to show that

(***) tan(AOB/2) > tan(CTD/2).

From triangle AOB we get that

tan(AOB/2) = AB/(2*R

_{1})

tan(CTD/2) = CD/(2*R

_{2})

Since AB=CD, and R

_{1}<We obtain (***), and therefore (*).

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