Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
This entry contributed by Peter Tran.
Click the figure below to view more details of problem 1213.
Wednesday, May 11, 2016
Geometry Problem 1213: Triangle, Circumcircle, Altitudes, Cyclic Quadrilateral, Congruence
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ACDF is a cyclic quadrilateral and DF extends to N.
ReplyDeleteSo angle AFN = C.
ACBM is a cyclic quadrilateral and BM extends to M.
So angle AMN = C.
It follows AFMN is a cyclic quadrilateral.
Next angle ANM
= sum of angles MNF and FMA
= sum of angles MAF and AMF
= angle AFE = C
(last equality true since quad BFEC is cyclic).
So Angles ANM and ANM are equal (each being C).
Hence AM = AN.
Let H be the Orthocentre
ReplyDelete< BCE = C
= < DHC (DCEH being cyclic)
= < AFD (BFHD being cyclic)
= < AFN (alternate angles)….(1)
But < AFE = C (AFEC being cyclic)….(2)
And < AMN = C (exterior angle)….(3)
From (1) and (3), AFMN is cylic
Hence < ANM = C (exterior angle of cyclic quad AFMN)…(4)
From (3) and (4), AM = AN
Sumith Peiris
Moratuwa
Sri Lanka