Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 866.
Saturday, March 30, 2013
Problem 866: Cyclic Quadrilateral, Circle, Rectangle, Center, Congruence, 90 Degrees
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ReplyDeleteconstruct a rectangle AMND (with center L) that AM=BC.
Angle ABC+ angle CDA=180. Because angle EBA+angle HBC=90+90=180 degrees ,is ,angle EBH+angle ABC=180.So, angle EBH=angleCDA . Additional, angle HBG=angle ADL and angle EBF=angle CDK.So, angle O1BO2=angle LDO3. The triangles O1BO2, LDO3 have additional,O1B=DO3 ,DL=BO2 ,therefore are equal ,so, O1O2=LO3 (1)
Angle BCD+ angle DAB=180. Because angle BCG+angle KCD=90+90=180 degrees ,is ,angleBCD +angle KCG=180 degrees . So, angle KCG=angleBAD . Additional, angle HCG=angle LAD and angle KCO3 =angleO1AB .So, angle O1AL=angle O2C O3. The triangles O1AL 2,
O2C O3 have additional,O1A=CO3 ,AL=CO2 ,therefore are equal ,so, O2O3=LO1 (2)
From (1),(2), O1O2O3L it is a parallelogram.
because triangle O1O2B = triangle LDO3 ,is , angle LO3D = angle O2O1B .But , angle EO1B= angle CO3D.Therefore ,angle EO1B-angle O2O1B= angle CO3D- angle LO3D ,so,
angle EO1O2=angle CO3L. Additional, angle CO3O2 =angle AO1L.
angle EO1O2+angle O2O1L+angle AO1L=180 ,therefore, 2angle O2O1L=180 degrees , angle O2O1L=90 degrees.So , O1O2O3L is, rectangle ,therefore angle O1O2O3L=90 degrees. Ο.ε.δ