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

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

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## Thursday, October 8, 2015

### Geometry Problem 1151: Circle, Diameter, Chord, Perpendiculars, 90 Degrees, Metric Relations

Labels:
chord,
circle,
diameter,
metric relations,
perpendicular

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Let the radius be R and let OE = s

ReplyDeleteLet AG = a. GF = b and GH = c

From similar Tr.s

(R+s)(R-s) = CE^2 = ED^2. (1)

(R+s)(2R) = a(a+b). (2)

(R+s)^2 = a(a-c). (3)

(2) - (3)

(R+s)(R-s) = a(a+b) - a(a-c) = a(b+c) which from (1) equals CE ^2 which is the required result.

Sumith Peiris

Moratuwa

Sri Lanka

Since BEGF concyclic, thus

ReplyDeleteCE² = AE×EB

= AE×AB - AE²

= AG×AF - AG×AH

= AG×HF

For (1) I've used similar Tr.s AED & BED, for (2) Tr.s ABF and AEG and for (3) Tr.s AEH & AEG

ReplyDelete