Geometry Problem. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Friday, October 5, 2018

### Geometry Problem 1392: Two Cyclic Quadrilateral, Cyclic Octagon, Circle

Labels:
circle,
concyclic,
cyclic quadrilateral,
geometry problem,
octagon

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ang: MEF + PIJ = 180, MLK + GHP = 180 => AEML, PICH cyclic

ReplyDelete=> EML + HPI = 180

Let the angles of Tr. FGB be a,b,x respectively

ReplyDeleteSimilarly for FGN be c,d,u

For ELA be d,e,y

For LEM be a,f,w

For JKD be g,f, 180-x

For JKQ be e,h,v

For IHP be b,g, z and

For HIC be c,h, 180 - y

Write 8 sum of angles equations for the 8 Tr.s each adding to 180.

Eliminate a,b,c,,d,e, f,g, h from these 8 equations

We end up with u + v = w + z

So each side must be = to 180 which is our desired result

Sumith Peiris

Moratuwa

Sri Lanka