The side BC of $∆$ ABC is produced to D, the bisector of $∠$ A meets BC at L, Prove that $∠$ ABC + $∠$ ACD = 2 $∠$ ALC.

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Solution

$Let ∠ BAL = ∠ LAC = x and ∠ ABC = y In △ ABC, we have : 2 x + y + ∠ ACB = 180 ° \Rightarrow ∠ ACB = 180 ° - 2 x - y . . . (i) Also, ∠ ACD = ∠ BAC + ∠ ABC = 2 x + y (Exterior angle property) \Rightarrow ∠ ACD = 2 x + y . . . (ii) And in △ ALC, we have : x + (180 ° - 2 x - y) + ∠ ALC = 180 ° \Rightarrow - x - y = - ∠ ALC \Rightarrow ∠ ALC = x + y . . . . (iii) Now, RHS : 2 ∠ ALC = 2 (x + y) and LHS : ∠ ABC + ∠ ACD = y + 2 x + y = 2 (x + y) ∴ LHS = RHS Hence, proved .$

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