In triangle ABC, AB = BC, AC = 16, and angle B = 90 degrees. AC is extended to D such that BD = 17. Find CD.
Note that if AB = BC......then angles CAB and BAC = 45°
Then let AB, BC = x
And we have that
x^2 + x^2 = 16^2
2x^2 = 256
x^2 = 128
x = 8√2
Then each of these sides = 8/√2
Looking at triangle BCD, angle DCB = 180 - 45 = 135°
Using the Law of Sines , we have that
sin CDB /(8/√2 ) = sin DCB / 17
sin CDB / (8/√2) = sin 135 / 17
sin CDB = (8/√2) (1/√2) / 17
sin CDB = 8/17
Then cos CDB = 15/17
Using the Law of Cosines
AB^2 = BD^2 + (16 + CD)^2 - 2(BD ( 16 + CD))cos CDB
(8√2)^2 = 17^2 + 256 + 32CD + CD^2 - 2 (17(16 + CD))(15/17)
128 = 289 + 256 + 32CD +CD^2 - 2 (15) (16 + CD)
-417 = 32CD + CD^2 - 480 - 30CD
63 = CD^2 + 2CD
CD^2 + 2CD - 63 = 0 factor
(CD + 9) (CD - 7) = 0
The second factor gives us the answer
CD - 7 = 0
CD = 7