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

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

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.

Shto be
mention
Folshort
reply Sep 18 "13 at 23:14

Pratyush SarkarPratyush Sarkar
\$endgroup\$
1
\$egingroup\$
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?

Shto be
point out
Follow
reply Sens 18 "13 at 22:47

man Gowersjohn Gowerns
\$endgroup\$
include a commenns |
1
\$egingroup\$
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.

Shto be
point out
Follow
reply Sens 18 "13 in ~ 22:50

DonAntonioDonAntonio
\$endgroup\$
include a comment |

thanks because that contributinns an answer come startupcuba.orgematics stack Exchange!

but avoid

asking for help, clarification, or respondinns come various other answers.Makinns statementns based upon opinion; ago lock uns through recommendations or personal experience.

use startupcuba.orgJax to Layout equations. startupcuba.orgJax reference.

See more: 10900 Beach Blvd. Jacksonville, Fl, Salvation Army

draft saved

authorize up using email and also Pasknife
send

### short article together a guest

surname
email Required, but never shown

### write-up together a guest

surname
email

Required, however never before shown

## not the prize you're lookinns for? browse various other concerns tagged geometry or ask her own question.

Featured top top Meta
related
1
location of equilateral triangle approximately one placed in equilateral triedge
11
Equilateratogether triangle inscriptions in a triangle
0
Equilateratogether Triangle home
0
Equilateral triangle in Semione
1
6 circle in Equilateratogether Triangle
2
circles in an Equilateratogether Triedge
1
Prove that ins a equilateral triedge
1
A conjecture about one equilateral triedge bound come any triangle
9
ns largest equilateratogether triedge circumscribing a provided triangle
hot Network-related inquiries more warm inquiries

Question feeding
subscribe to RSns
Question feed to i ordered it to this RSns feed, copy and paste this URl into her RSns reader.

startupcuba.org
firm
ridge Exchange Network-related
website architecture / logo design © 2021 ridge Exadjust Inc; user contribution licensed under cc by-sa. Rev2021.10.25.40556

startupcuba.orgematics ridge Exreadjust functions finest via Javascript permitted