Just display that triangle \$AOB, BOC, COA\$ room all congruent (then all the political parties of triangle \$ABC\$ have to be equal).

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.

re-superstructure
cite
monitor
answer Sep 18 "13 at 23:14

Pratyush SarkarPratyush Sarkar
3,5401616 silver- badges3131 bronze title
\$endgroup\$
include a comment |
1
\$egingroup\$
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?

re-publishing
point out
follow
answered Sep 18 "13 in ~ 22:47

man GowersJohn Gowers
22.9k44 yellow badges5757 silver badges9696 bronze title
\$endgroup\$
add a comment |
1
\$egingroup\$
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.

share
point out
monitor
answer Sep 18 "13 in ~ 22:50

DonAntonioDonAntonio
203k1717 yellow badges116116 silver- badges269269 bronze title
\$endgroup\$
add a comment |

Thanks because that contributing response to inter-base.net Stack Exchange!

Please be certain to answer the question. Carry out details and share her research!

But avoid

Asking for help, clarification, or responding to various other answers.Making statements based upon opinion; ago them up with references or an individual experience.

Use inter-base.netJax to layout equations. inter-base.netJax reference.

To find out more, view our tips on writing good answers.

See more: 7 Days To Die Crossbow Vs Bow, Brazmock Archery Mod At 7 Days To Die Nexus

Draft saved

authorize up utilizing Google
send

Post as a guest

name
email Required, however never shown

Post together a guest

name
email

Required, but never shown

Not the prize you're feather for? Browse various other questions tagged geometry or questioning your own question.

Featured ~ above Meta
associated
1
Area of equilateral triangle around circles put in it is intended triangle
11
it is provided triangle enrolled in a triangle
0
it is provided Triangle residential or commercial property
0
equilateral triangle in Semicircular
1
6 one in it is provided Triangle
2
circles in an equilateral Triangle
1
Prove that is a it is intended triangle
1
A conjecture around an it is intended triangle bound to any kind of triangle
9
The biggest equilateral triangle circumscribing a given triangle
warm Network questions much more hot questions

inquiry feed
i ordered it to RSS
question feed To i ordered it to this RSS feed, copy and also paste this URL into your RSS reader.

inter-base.net
firm
ridge Exchange Network
site design / logo design © 2021 ridge Exchange Inc; user contributions licensed under cc by-sa. Rev2021.11.5.40661

inter-base.netematics ridge Exchange works ideal with JavaScript allowed

her privacy

By click “Accept every cookies”, you agree stack Exchange deserve to store cookies on your device and disclose info in accordance v our Cookie Policy.