Proposed Problem

Click the figure below to see the complete problem 355 about Circles, Common external tangent, Concyclic points.

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Complete Problem 355

Level: High School, SAT Prep, College geometry

## Sunday, September 13, 2009

### Problem 355: Circles, Common external tangent, Concyclic points

Labels:
circle,
common tangent,
concyclic,
cyclic quadrilateral

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join H to D, E to A

ReplyDeleteang CDN = ang NHD, ang NDH = ang GHN (same arc ND, NH)

=> ang D = ang H (1)

at the same way

ang A = ang E (2)

(1) & (2) => DH//AE

ang PNG = ang CND, ang QNC = GHN ( vertical ang )

ang BMQ = ang MEF, ang FMP = BAM

=> ang BMF + ang GNC = BMQ + M + FMP + QNC + N + PNG

=> (BMQ + FMP) + M + (QNC + PNG) + N

=> (BMQ + FMP + QNC + PNG) + M + N

=> ( A + D ) + M + N = 360

=> 180 + M + N = 360 ( A + D = 180, DH//AE )

=> M + N = 180

Let < AEM = < CAM = < AMB = < FMP = p

ReplyDeleteLet < FME = < FEM = q

Let < GNH = < GHN = < QNC = < HND = r

Let < CND = < CDN = < DHN = s

Let HE and DA be extended and meet at X. Then AEX and DHX are both isosceles triangles. Hence ADHE is an isosceles trapezoid.

Hence < AEH + < ADH = (p+q) + (r+s) = 180

But < EMP = p+q and < QND = r+s

So < EMP + < QND = 180

Therefore < QMP + QNP = 180

Hence M,P,N,Q are concyclic.

Sumith Peiris

Moratuwa

Sri Lanka