Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 814.

## Monday, October 15, 2012

### Problem 814: Two Equilateral Triangles with a common Vertex, Angle

Labels:
60 degrees,
angle,
equilateral,
triangle,
vertex

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Observe that triangle BCF is congruent to triangle BAD by rotation.

ReplyDeleteLet angle BCF = x, then angle BAD = x,

Angle AGF = angle ACF - angle CAG = (60+x)-(x-60)=120

@W Fung: The proof is forced. We do not know anything about angles FBC and DBA or sides (CF) and (AD) so that we can say that triangle (BAD) is obtained from the triangle (BCF) using rotation.

ReplyDeleteTo Rusu:

DeleteI skipped the short proof of congruence as i believe it is too obvious.

For your information,

BA = BC ; BD = BF ; angle ABD = 60 - angle DBC = angle FBC

So by S-A-S, triangle ABD and CBF are congruent. Hence the rotation at point B exists.

Rotation clockwise 60° about B.

ReplyDeleteB→B

C→A

F→D

ΔBCF→ΔBAD

∴ΔBCF≡ΔBAD

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ΔBCF≡ΔBAD

⇒ ∠BCF = ∠BAD

⇒ ∠BCF = ∠BAG

⇒ B,A,G,C concyclic

⇒ ∠AGC = 180° - ∠ABC = 120°

⇒ ∠AGF = 120°

solution to the problem 814( without rotation), by Michael Tsourakakis from Greece

ReplyDeleteangle x+angle ω=60 and angle y+angle ω=60, so,angle x=angle y

The triangles ABD, BCF ,are equal ,why have, AB=BC ,BD=BF and angle x=angle y Therefore,angle BDA=angle BFC .So, the quadrilateral BCDF ,is , cyclic quadrilateral. Therefore,angleFGD=angleDBF=60. So, angle AGF=120

Image: http://img4.imageshack.us/img4/4129/56463515.png

http://www.mathematica.gr/forum/viewtopic.php?f=20&t=31981&p=148019

ReplyDelete< ABD = < CBF so Tr.s ABD and CBF are congruent SAS

ReplyDeleteHence < ADB = < BFG .= say @

So < AGF = < ADF + DFG = @+60 + 60-@ = 120

Sumith Peiris

Moratuwa

Sri Lanka

Further it follows that BGDF and ABCG are cyclic

ReplyDelete