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

Labels:
altitude,
circumcircle,
congruence,
cyclic quadrilateral,
triangle

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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