In triangle ABC, AB = 14, AC = 13, and BC = 15. Let X be the reflection of A over B, and let CX intersect the circumcircle of triangle ABC again at D. Find the length CD.
+128472 [ Because of the nasty roots that we will find in this problem....I leaned heavily on WolframAlpha for most of the computations ]
Note that AB = BX so.....AX = 28
Using the Law of Cosines we have
BC^2 = AB^2 + AC^2 - 2(AC*AB) cos CAX
15^2 = 14^2 +13^2 - 2(13 * 14) cos CAX
Solving for the cos CAX we get that cos (CAX) = 5/13
So....the sin of CAX = 12/13
And we can find CX using the Law of Cosines again
CX^2 = AC^2 + AX^2 - 2(AC * AX) cos CAX
CX^2 = 13^2 + 28^2 - 2(13 * 28) (5/13)
Solving this for CX gives us CX = √673
And again we can find the cosine of angle BCX as
BX^2 = BC^2 + CX^2 - 2(BC * CX) cos BCX
14^2 = 15^2 + 673 -2 ( 15 * √673) cosBCX
Solving this for cos BCX gives us 117/ [ 5 √673 ]
And sin BCX = √ [ 1 - [ 117/ [ 5 √673 ] ]^2 ] = 56 / [ 5√673]
Connect BD
Since quadrilateral ABDC is inscribed in a circle....angles CAB and CDB are supplemental....so the have the same sines =12/13
So.....using the Law of Sines
BC / sin CDB = BD /sin BCX
15 / (12/13) = BD / [ 56 / [ 5√673] ]
Solving this gives that BD = 182 / √673
Finally.....using the Law of Cosines we can find CD as folllows
BD^2 = BC^2 + CD^2 - 2 (BC * CD) cos BCX
Let CD = x and we have that
182^2 / 673 = 15^2 + x^2 - 2 (15 * x)*(117/ [5sqrt (673)] )
Solving this for x = CD gives us two values
CD = 281 / √673 ≈ 10.83 or
CD = 421 / √673 ≈ 16.23
But BC = 15 and is clearly greater than CD
So CD = 281 / √673 ≈ 10.83 is the correct solution
