 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Equal Tangents (Posted on 2010-10-03) C(UV) denotes the circle with diameter UV.

T(P,QR) denotes the tangential distance |PS|,
where point P lies outside C(QR), point S
lies on C(QR), and PS is tangent to C(QR).

Let A, B, C, and D be distinct, collinear
points in that order.

Construct a point E on line AD such that
|EF| = T(E,AB) = T(E,CD) = |EG|

 See The Solution Submitted by Bractals Rating: 3.0000 (1 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re: Algebraic solution leads to geometric | Comment 10 of 11 | (In reply to Algebraic solution by Larry)

Taking the algebraic solution, I solved for [C-E] and [E-B] to get an idea of the relative position of E between B and C.

Consider an x axis where A is at 0, and B,E,C, and D are at x values of B,E,C, and D.
E = (D*C)/(D+C-B)
C-E=  ( C )* {(C-B) / (D+C-B)}
E-B= (D-B) * {(C-B) / (D+C-B)}
So, the ratio of the two line segments |BE| and |EC| is (D-B)/C.

So for a geometric construction, draw any line "L1" (except horizontally) through C.  Mark onto "line L1" point "M" which is distance C away from point C  (the length C is the distance between A and C).

Now start from point "M", continue along line "L1" in the same direction and mark out the length D-B and call it point "N".  Now draw line "L2" from point N to point B.  Draw line "L3", parallel to "L2" but passing through point M.  Where line "L3" intersects line ABCD is going to be point E.  By using similar triangles we have divided line segment |BC| into the proper ratio.

Edited on October 4, 2010, 10:36 pm
 Posted by Larry on 2010-10-04 22:34:19 Please log in:

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