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

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

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## Saturday, February 14, 2015

### Geometry Problem 1083: Semicircle, Diameter, Perpendicular, 90 Degrees, Tangent Line, Metric Relations

Labels:
90,
diameter,
metric relations,
perpendicular,
semicircle,
tangent

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through D draw line perpendicular to AB, end on F;

ReplyDeleteConnect GC,

triangle GDF similar to triangle DGE, ( angle AGD = angle GCD)

a is radius of circle

so X^2 = 2a(a+b)

http://s29.postimg.org/m7r8fv3hz/pro_1083.png

ReplyDeleteDraw lines and points per sketch

Lot O is the center of the circle, we have FO⊥AE and GO⊥AB => AFOG is a square

So CD= 2.AF= 2a

We also have ∠ (HDG) =∠ (BGC) and ∠ (CDG)= ∠ (BGC)

so GD is an angle bisector of angle CDH

DH=GK= a + b

Relation in the right triangle DGC give DG^2= DK.DC

or x^2=2a.(a+b)

May I point out a typo ?

ReplyDeleteInstead of DH=GK= a + b, should it not be DH=DK= a + b ?

Thanks for the correction.

ReplyDeletethe corrected statement is " DH=DK=a+b"

Peter Tran

https://www.dropbox.com/s/uggeh3qzew0fdld/1083.fig?dl=0

ReplyDeleteComplete the rectangle BAEJ and semicircle to full circle.

Let EJ and diameter GOK intersect at H.

Then x^2 = GD^2 = GK.GH = 2a(a + b).

Let H be on AG such that HD//BC//GO where O is the centre of the semicircle CGD

ReplyDeleteAGOF is a square of side a and the radius of the semicircle is a

Tr. a CDG and GDH are similar hence

2a/x = x/(a+b) from which the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Further if DE = c and DF = y then it can be shown that

ReplyDeletey^2 = 2ac

Moreover if J is the food of the perpendicular from G to CD,

ReplyDeleteCJ = a-b and GJ^2 = a^2 - b^2