hi polythenepam,
PQ and RS are common tangents to two intersecting circles. IF T is one of the points of intersection of the two circles, prove that the circles through T,P,Q and through T,R,S touch eachother
since both circles TPQ and TRS pass through T, then if they ONLY touch eachother, it would have to be at T. in other words, both circles TPQ and TRS share a
common tangent through T.
since T can be either intersection points of the original two circles, then for clarity and unambiguity in my proof below, i will need to fix it to one point. so, let T be the intersection point that is, say, on the side
closest to the common tangent RS.
now, draw a tangent at T to the new circle TRS. Produce (extend) this tangent so that it cuts the original circle with P on it at X, and cuts the original circle with Q on it at Y. (ie. the original circles are now PTX and QTY)
okay...now for the notation and variables:
let "<" denote "angle"; let
< TPQ= x , < TQP= y, and < RTS= z
now for the proof:
< PQT = y = < QYT (alt. < in opposite segment theorem, tangent being PQ)
< TRS = < STY (alt. < in opposite segment theorem, tangent being XY)
but, < STY = < YSS' ,
where S' is any point on RS away from S in the direction R-> S, (alt. < in opposite segment theorem, tangent being RS)
therefore, < TRS = < YSS' ---> TR//YS
hence,
< RTS = z = < TSY (alt < 's on //lines)
and so, < TQY = 180 - < TSY = 180 - z (opposite < 's in cyclic quad. TQYS are supplementary)
< QTY = 180 - < TQY - < QYT (< sum of triangle TQY) = 180 - (180-z) -y =
z - y
and in similar fashion as above, you get:
< PTX = z - x
now, < PTQ = 180 - < QPT - < PQT = 180 -x -y
and, < PTQ + < PTX + < QTY = 180 (< sum of straight line).
ie. 180 = (180 -x -y) + (z -x) + (z-y) --->
z = x + y
therefore, < PTX = z - x = y , and < QTY = z - y = x ;
hence,
< PTX = y = < PQT, and
< QTY = x = < QPT
----->(through the alt. < in opposite segment theorem)
XY MUST be tangent at T to circle TPQ.
hence, circles TRS and TPQ share a common tangent XY at T, which means they touch at T.
hope that helps
