## Wednesday, May 11, 2016

### Geometry Problem 1213: Triangle, Circumcircle, Altitudes, Cyclic Quadrilateral, Congruence

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. 1. ACDF is a cyclic quadrilateral and DF extends to N.
So 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.

2. Let H be the Orthocentre

< 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