Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below for more details.

## Wednesday, October 7, 2015

### Geometry Problem 1150: Circle, Perpendicular Secants, 90 Degrees, Congruence

Subscribe to:
Post Comments (Atom)

Extend AH to M which is on BE. < MEA = < ACD = < EAM = < GCA = < DCM since Tr. ACD & MCD are therefore congruent. So 2AH = AM = ME = BM = AG

ReplyDeleteNow I need to show that BM = GF or that GB = FM which step eludes me still. Dear Antonio could u help?

Sumith Peiris

Moratuwa

Sri Lanka

Am I on the right track Antonio?

DeleteDear Sumith,

DeleteIn your first statement < MEA = < ACD = < EAM = < GCA = < DCM

why <DCM = < GCA = ....

Thanks

Antonio

Extend CG and EA let be P point where them meet each other. Draw MN //GF9Through A). Triangles PMA kongruent to AHD(PCD isoceles) PMA kongruent to ADN => MA=AN=AH

ReplyDeleteExcellent!

DeleteCBDE is cyclic, so <BEA=<ACD (1), CGAE is cyclic, so <BEA=<BCG (2), hence BC is bisector of <GCD (3). Let M be the intersection of AD and CG, N the intersection of BC and DF; from (3) and AD_|_BC, CMND is a rhombus. GF is one of its altitudes, and AH is half of another altitude, done.

ReplyDeleteBecause CBDE is cyclic, a=<BED=<BCD

ReplyDeleteBecause CGAE is cyclic, <BED=<GCB=a

Therefore, <GCD=2a.

GF=CD*sin(2a)=2CDcos(a)sin(a)=2cos(a)(CDsin(a))=2cos(a)AD.

Note that from similar triangles, <HAD=<ACD=a, so ADcos(a)=AH.

Plugging this into previous equation gives GF=2cos(a)AD=2AH.