## Saturday, February 14, 2015

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

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it. 1. through D draw line perpendicular to AB, end on F;
Connect GC,
triangle GDF similar to triangle DGE, ( angle AGD = angle GCD)
a is radius of circle
so X^2 = 2a(a+b)

2. http://s29.postimg.org/m7r8fv3hz/pro_1083.png
Draw 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)

3. May I point out a typo ?
Instead of DH=GK= a + b, should it not be DH=DK= a + b ?

4. Thanks for the correction.
the corrected statement is " DH=DK=a+b"

Peter Tran

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

Complete 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).

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

AGOF 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

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

y^2 = 2ac

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

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