D lies inside the triangle ABC. ∠BAC = 50o. ∠DAB = 10o, ∠DCA = 30o, ∠DBA = 20o. Show that ∠DBC = 60o.
Solution
Reflect A in the line BD to get A'. Let Z be the intersection of BD and AA'. Let BA' meet AC at X. Since ∠ABX = 2 ∠ABD = 40o, and ∠BAX = 50o, we have ∠BXA = 90o. Now ∠DAA' = ∠BAA' - ∠DAB = ∠BAA' - 10o. But ∠BAA' = 90o - ∠DBA = 70o, so ∠DAA' = 60o.
Let BX meet CD at Y. ∠DYX = ∠YXC + ∠DCX = 90o + 30o = 120o = 180o - angle DAA', so DAA'Y is cyclic, so ∠A'YA = ∠A'DA = 2 ∠ZDA = 2(∠DBA + ∠DAB) = 60o.
But ∠XYC = 90 - ∠DCA = 60o, so C is the reflection of A in BX. Hence BC = BA, so ∠ACB = ∠BAC = 50o. Hence ∠ABC = 80o and ∠DBC = 80o - ∠DBA = 60o.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002