I give one, should be a poor one. let Angle CAD be a1, CBD a2, ACD a3, ABD a4, BCD a5, BAD a6 as to triangle PQR, angle PQS be b1, QPS b2, RQS b3, QRS b4, RPS b5, PRS b6 Using CD * sin 120 = b * sin a1 = a * sin a2 AD * sin 120 = b * sin a3 = c * sin a4www.ddhw.com BD * sin 120 = a * sin a5 = c * sin a6 considering triangle PQR, we have b * sin b1 = a * sin b2 b * sin b3 = c * sin b4 a * sin b5 = C * sin b6 together with 60 = a1 + a3 = a2 + a5 = a4 + a6 60 = b1 + b3 = b2 + b5 = b4 + b6 thus all the 12 angles are less than 90 degrees. we must have a1 = b1, a2 = b2, and so on. If not, www.ddhw.com suppose a1> b1, then a2 > b2, since b * sin a1 = a * sin a2 and b * sin b1 = a * sin b2 b3 > a3 (since a1 + a3 = b1 + b3, and a1 > b1) b5 > a5 (since a2 + a5 = b2 = b5, and a2 > b2) b3 > a3 gives b4 > a4 (since b * sin a3 = c * sin a4, and b * sin b3 = c * sin b4) b5 > a5 gives b6 > a6 but b4 + b4 = a4 + a6 = 60. contradiction. so a1 = b1 thus proved www.ddhw.com
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