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

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

Details: Click on the figure below.

## Sunday, January 12, 2020

### Geometry Problem 1453: Two Semicircles, Cyclic Quadrilateral, Concyclic Points

Labels:
circle,
concyclic,
cyclic quadrilateral,
diameter,
semicircle

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Let:

ReplyDelete1.)

∠EOB = α

∠EQO = β

=> ∠OEQ = 180-α-β

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2.)

∠OBE = ∠OEB = 90-α/2

∠QCE = ∠QEC = 90-β/2

=> ∠BEC = α/2 + β/2

____________

3.)

∠EOF = γ (given)

=> ∠EQG = 360-2α-2β-γ

____________

4.)

We know:

∠EQC + ∠EQG = β+360-2α-2β-γ = 360-2α-β-γ = ∠GQC

=> ∠QCG = 90-∠GQC/2 = α+β/2+γ/2-90 = ∠HCB

____________

5.)

We know:

∠EOQ + ∠EOF = α+γ = ∠FOB

=> ∠O(C)BF = 90-∠FOB/2 = 90-α/2-γ/2 = ∠HBC

____________

6.)

From (4) and (5):

∠CHB = ∠BHC = ∠FHG = ∠GHF = ∠MHN = 180-α/2-β/2

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7.)

∠ECB-∠HCB = ∠MCH = α/2 + β/2 + γ/2

=> 180 - ∠CMH = ∠EMH = 180 - α/2 - β/2 - γ/2

____________

8.)

∠EBC-∠HBC = ∠NBH = γ/2

=> ∠MHC = ∠NHB = 180 - ∠CHB = α/2 + β/2

=> 180 - ∠BNH = ∠ENH = α/2 + β/2 + γ/2

____________

For a Cyclic Quadrilateral is:

(∠MEN = ∠BEC) + ∠MHN = ∠EMH + ∠ENH = 180

https://photos.app.goo.gl/eupasLBXW1f4WeJB9

ReplyDeleteDraw altitude EP

Let ∠ (BEP)= u and ∠ (PEC)=v

Note that AEB and CED are right triangles

In right triangle AEB we have ∠ (PAE)= ∠ (PEB)= u

In circle O ∠ (PAE)= ∠ (PFE)=u

Similarly ∠ (PEC)= ∠ (CDE)= ∠ (CGE)=v

In triangle FHG angle( u+v) supplement to angle ∠ (FHG)

So ∠ (CEB)=u+v supplement to ∠ (FHG) => ENHM is cyclic quad.

Let

ReplyDelete< CEB = u

< FEA = < FBA = v

< GED = < GCD = w

So < BAF = 90 - v = < BEG = 90 - u + w and so u = v + w

This yields < CHM = < MEN and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Let ∠FAB=x, ∠GDC=y

ReplyDeleteHence ∠FBA=90-x and ∠GCD=90-y => ∠CHB=MHN=x+y -----(1)

Since AFEB is concyclic, ∠FEB=180-x=>∠GEB=x

Similarly CEGD is concyclic, ∠CEG=180-y=>∠FEC=y

=>∠CEB=180-(x+y) which is suppliment to (1)