Geometry Problem

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

Click the figure below to see the complete problem 639.

## Saturday, July 23, 2011

### Problem 639: Semicircle, Diameter, Perpendicular, Inscribed Circle, Tangent, Arbelos, Congruence

Labels:
arbelos,
circle,
congruence,
diameter,
inscribed,
perpendicular,
semicircle,
tangent

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Connect AG

ReplyDeletePer the result of problem 636, we have AG perpen. To DH.

Triangle CDH congruence to triangle GDA ( Case ASA)

So AD= DH

Peter Tran

Let AC = 2a, BC = 2b, FG = c and FH = x

ReplyDeleteThen from similar Tr.s we can show that x = c(b+c)/(b-c)

Applying Pythagoras twice to Tr. CDF,

(a+b-c)^2 - (a+c-b)^2 = (b+c)^2 - (b-c)^2 which simplifies to a = bc/(b-c)

So DH = x+b+c = b(b+c)/(b-c)

DA = 2a+b which is again = to b(b+c)/(b-c) by substituting a = bc/(b-c)

So DA = DH

Sumith Peiris

Moratuwa

Sri Lanka

Some clarifications on my above proof

ReplyDeleteIf M is the foot of the perpendicular from F to AB then we can show that OM = b+c-a. This implies that OD = a

Above I have written the Pythagorean expressions twice for FM