You are watching: How can you prove a triangle is an equilateral triangle?

To show this, note that triangle $A"BO, A"OC$ are equilateral due to the fact that they are all radius the the circles. Heat $OA"$ is perpendicular to line $BC$. Hence, $angle OBC = angle OCB = 30^circ$. For this reason $angle BOA" = angle COA" = 60^circ$. Now, $angle BOA = 180^circ - angle BOA" = 180^circ - 60^circ = 120^circ$ because $A"A$ is a directly line. Similarly, $angle AOC = 120^circ$. For this reason $angle BOC = angle BOA = angle AOC = 120^circ$. Currently its straightforward to display triangles $AOB, BOC, COA$ space all congruent due to the fact that they space isosceles and have typical equal sides.

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answer Sep 18 "13 at 23:14

Pratyush SarkarPratyush Sarkar

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**Hint:**sign up with the 2 dots (the points where the horizontal line intersects the 2 circles) to the suggest just over them where the two circles intersect. What deserve to you say around the triangle you have just formed?

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answered Sep 18 "13 in ~ 22:47

man GowersJohn Gowers

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Call the rightmost peak of the triangle $;A;$ , the top one $;B;$ and the lower one $;C;$, let $;M,N;$ it is in the left (right) circle"s center and let $;P;$ it is in the intersection point of $;AM;$ through $;BC;$. Be sure you have the right to prove (or at the very least follow) the following:

=== $;MNperp BC;$ (the center"s segment is constantly perpendicular come the usual cord of 2 intersecting, non-tangent, circles)

=== Thus, $;BP=PC;$ (a straight segment v a circle"s facility is perpendicular to a cord iff the bisects it)

=== in $;Delta ABC;$ , we have actually that $;APperp BC;$ is also the mean to $;BC;$ and also thus $;Delta ABC;$ is isosceles, through $;AB=AC;$

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

=== $;Delta NBM;$ is equilateral and also since $;angle BNM;$ is a main angle in the best circle and it equals... Then...finish the exercise.

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answer Sep 18 "13 in ~ 22:50

DonAntonioDonAntonio

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