just present the triangles $AOB, BOC, COA$ to be every congruent (climate every the political parties the triedge $ABC$ have to be equal).

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to display this, Keep in mind that triangles $A"BO, A"OC$ are equilateral Since castle are every radius of ns circles. Line $OA"$ ins perpendicular to line $BC$. Hence, $edge OBC = edge OCB = 30^circ$. So $edge BOA" = angle COA" = 60^circ$. Now, $angle BOA = 180^circ - edge BOA" = 180^circ - 60^circ = 120^circ$ Because $A"A$ is a straight line. Similarly, $edge AOC = 120^circ$. Therefore $angle BOC = edge BOA = angle AOC = 120^circ$. Currently itns straightforward to show triangle $AOB, BOC, COA$ to be all congruent Since castle are isosceles and also have Typical equal sides.


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Pratyush SarkarPratyush Sarkar
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Hint: sign up with ns two dotns (ns pointns where ns horizontal line intersectns the 2 circles) come ns point simply over them where the 2 circles intersect. Wcap have the right to girlfriend speak about the triedge girlfriend have actually simply formed?


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man Gowersjohn Gowerns
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speak to ns rightMost peak that ns triangle $;A;$ , ns top a $;B;$ and the reduced one $;C;$, lens $;M,N;$ be the lefns (right) circle"s center and also lens $;P;$ it is in the interarea suggest that $;AM;$ through $;BC;$. Be certain friend can prove (or at leastern follow) the following:

=== $;MNperp BC;$ (the center"ns segment ins always perpendicular to the Typical cord of 2 intersecting, non-tangent, circles)

=== Thus, $;BP=PC;$ (a straight segment with a circle"ns facility is perpendicular come a cord iff ins bisects it)

=== in $;Delta ABC;$ , us have actually the $;APperns BC;$ is additionally the Typical come $;BC;$ and thus $;Delta ABC;$ ins isosceles, with $;AB=AC;$

=== $;BMCN;$ is a rombus, and also therefore $;BM=MC=$radius

=== $;Delta NBM;$ ins equilateratogether and also Due to the fact that $;angle BNM;$ ins a main edge in ns right circle and ins equals... Then...complete ns exercise.


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DonAntonioDonAntonio
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