A Complete Proof about Tangential Circles

Date: 06/05/98 at 06:32:01
From: Emma
Subject: Writing a proof

Dear Dr Rob,

Thank you for your previous help with writing a proof. I'm not sure 
how much justification to include in proofs, so I would like to see a 
proof with quite a bit of full justification. I would love to get to 
the end of this proof: 

   Two circles of the same radius touch at A and have a common tangent 
   BC. A small circle touches the larger circles and also has BC as a 
   tangent. Prove that the radius of the small circle is 1/4 of 
   the radius of the larger circles.

All I know is that the Pythagorean Theorem should be used with R and r 
as the radii. What would be the proof be if it is concise with full 
justification? Thank you for all your help. I appreciate it greatly. 

Date: 06/05/98 at 09:52:42
From: Doctor Rob
Subject: Re: Writing a proof

Emma,

Here is one with a lot of justification. For the parts you think are 
obvious you can delete the reasons (in parentheses). If you think any
of the steps can be combined, you can do that, too. If any of the 
reasons I have used are not ones you know to be true, you may have to 
prove them or else find them in a geometry book. If you know more 
powerful reasons that will allow you to skip over some steps, you can 
do that, too, if you supply the reason.

Here is a diagram that may help you follow the proof:

      

Proof:

Let B and C be the points of tangency of the line and the two large
circles. Let D be the center of the circle tangent at B, E be the 
center of the circle tangent at C, and F the center of the small 
circle. (Given.)

Connect A and D with a line segment. Connect A and E with a line 
segment. Connect A and F with a line segment. Connect F to E and F to 
D with line segments. Connect B to D and C to E with line segments.  
(There is a unique line through any two points.)

Let G be the intersection of DF with circles D and F. It is the point 
of tangency of the two circles. (The point of tangency of two tangent 
circles lies on the line connecting their centers.)

Let H be the intersection of EF with circles E and F. It is the point 
of tangency of the two circles. (Same reason.)

<DAE is a straight angle; that is, <DAE = 180 degrees. (Same reason.)

Let r be the radius of circle F, R the radius of circles B and C. Note 
that 0 < r < R. (Given.)

Then FD = FG + GD = r + R = FH + HE. (FG and FH are radii of circle F, 
GD is a radius of circle D, and HE is a radius of circle E.)

Also DA = R = EA. (DA is a radius of circle D and EA is a radius of 
circle E.)

AF = AF, so triangles AFD and AFE are congruent. (If two triangles 
have the three sides of one equal to the three sides of the other, 
they are congruent. [S.S.S.])

<FAD = <FAE. (Corresponding parts of congruent triangles are 
congruent.)

<FAD + <FAE = <DAE = 180 degrees. (The whole is equal to the sum of 
its parts.)

<FAD + <FAD = 180 degrees. (Substitution.)

2*<FAD = 180 degrees. (Property of 1, Distributive Law, 1 + 1 = 2.)

<FAD = 90 degrees. (Equals divided by equals are equal.)
   Comment: This means that triangle FAD is a right triangle. We know 
   two sides, and now we will find side AF.

<DBC = 90 degrees, <ECB = 90 degrees. (A line is tangent to a circle 
if and only if it is perpendicular to the circle radius at the point 
of tangency.)

BD || CE. (Two lines perpendicular to the same line [BC] are 
parallel.)

BD = R = CE.  (They are radii of equal circles D and E.)

BCDE is a parallelogram. (A quadrilateral with two opposite sides 
parallel and equal is a parallelogram.)

BCDE is a rectangle. (A parallelogram one of whose angles is a right 
angle is a rectangle.)

AF || CE || BD. (Lines perpendicular to the same line [DE] are 
parallel.)

Extend AF to meet BC at point I.  (A line can be extended indefinitely 
in either direction. Two lines which are not parallel meet in a unique 
point.)

ADBI and AECI are squares. (AE = EC = R = AD = DB, and all the angles 
of these two quadrilaterals are right angles.)

AI = R. (The sides of a square are all equal.)

R = AI = AF + FI. (The whole is equal to the sum of its parts.)

FI is perpendicular to BC. (AECI is a square.)

I is the point of tangency of circle F and line BC, and FI is a radius 
of circle F. (A line is tangent to a circle if and only if it is
perpendicular to the circle radius at the point of tangency.)

FI = r. (It is a radius of circle F.)

R = AF + r.  (Substitution.)

AF = R - r. (Equals subtracted from equals are equal.  Also R > r from
above, so this is positive.)

AD^2 + AF^2 = DF^2. (Pythagorean Theorem.)

R^2 + (R-r)^2 = (R+r)^2.  (Substitution.)

R^2 = 4*r*R. (Equals [-(R-r)^2] added to equals are equal, and then
simplifying using the rules of algebra.)

R/4 = r. (Equals divided by nonzero equals [4*R > 0] are equal, and 
then more simplifying.)

This was what we were trying to prove, so we are done.

-Doctor Rob,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/