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Level: Mathematics Education, High School, Honors Geometry, College.
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Sunday, January 12, 2020
Geometry Problem 1453: Two Semicircles, Cyclic Quadrilateral, Concyclic Points
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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
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3.)
∠EOF = γ (given)
=> ∠EQG = 360-2α-2β-γ
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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
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5.)
We know:
∠EOQ + ∠EOF = α+γ = ∠FOB
=> ∠O(C)BF = 90-∠FOB/2 = 90-α/2-γ/2 = ∠HBC
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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
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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)