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Parallel-tangents-to-a-circle-at-A-and-B-are-cut-in-the-points-C-and-D-by-a-tangent-to-the-circle-at-E-Prove-that-AD-BC-and-the-line-joining-the-middle-points-of-AE-and-BE-are-concurrent-




Question Number 19321 by ajfour last updated on 09/Aug/17
Parallel tangents to a circle at A  and B are cut in the points C and D  by a tangent to the circle at E.  Prove that AD, BC and the line  joining the middle points of AE  and BE are concurrent.
ParalleltangentstoacircleatAandBarecutinthepointsCandDbyatangenttothecircleatE.ProvethatAD,BCandthelinejoiningthemiddlepointsofAEandBEareconcurrent.
Commented by ajfour last updated on 09/Aug/17
Answered by ajfour last updated on 10/Aug/17
Commented by ajfour last updated on 10/Aug/17
Let B be origin.  A(0,2r)  . Let m=tan θ  C((r/m), 2r)    ;  D(mr,0)  eq. of AD :   y=2r−(2/m)x  eq. of BC:   y=2mx  I (α,β) lies on both, hence    β=2r−((2α)/m)=2αm  ⇒  α(m+(1/m))=r  or  α=((mr)/(1+m^2 ))        β=((2m^2 r)/(1+m^2 )) .  x_E =rsin 2θ =((2mr)/(1+m^2 ))         x_P =x_M =(x_E /2) = ((mr)/(1+m^2 )) = 𝛂 .
LetBbeorigin.A(0,2r).Letm=tanθC(rm,2r);D(mr,0)eq.ofAD:y=2r2mxeq.ofBC:y=2mxI(α,β)liesonboth,henceβ=2r2αm=2αmα(m+1m)=rorα=mr1+m2β=2m2r1+m2.xE=rsin2θ=2mr1+m2xP=xM=xE2=mr1+m2=α.

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