In △ A B C , I is the incenter of the triangle. A C + A I = B C and A B + B I = A C . Find the measure of ∠ A B C , to the nearest degree.
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Nice problem.Took me quite a long time to solve.Please post more such challenging geometry problems.
Awesome problem! Keep posting such problems :)
P r o d u c e C B t o E w h i c h D B = B E . L e t ∠ A C B = θ , ∠ C A E = ∠ C E A = 9 0 ° − 2 θ ∵ D i s i n c e n t r e . ∠ D A B + ∠ D B A = 9 0 ° − 2 θ ∠ A D B = 9 0 ° + 2 θ . ∠ A D B + ∠ A E B = 1 8 0 ° ∴ A , D , B , E i s c o n c y c l i c . D B = B E ⇒ D A B = B A E = D A C l e t D A B = α ⇒ C A E = C E A = 3 α . ∠ A C E = 1 8 0 ° − 6 α ⇒ A C D = D C B = 9 0 ° − 3 α A B C = B A E + A E B = 4 α P r o d u c e B A t o F w h e r e A F = D A ⇒ B F = B C ⇒ B F C = B C F = 9 0 − 2 α ∠ F C D = F C B − D C B = ( 9 0 ° − 2 α ) − ( 9 0 − 3 α ) = α F C A = 9 0 ° − 2 α − ( 1 8 0 ° − 6 α ) = 4 α − 9 0 ∠ D C F = ∠ D A B = α . ∵ F C D A i s c o n c y c l i c A F = A D ⇒ ∠ F C A = ∠ A C D ⇒ 4 α − 9 0 ° = 9 0 ° − 3 α ⇒ α = 2 5 . 7 1 4 . . . ∠ B A B = 2 α = 5 1 . 4 2 8 . . .
Nice solution
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A more precise answer should be 7 3 6 0 ° ≈ 5 1 . 4 3 ° .
Cut B C into B R and C R , where C R = C A . As A C + A I = B C , A I = B R .
△ A C R is isosceles, with C I as an angle bisector. Therefore, C I is also the perpendicular bisector of A R , meaning that A I = R I , m ∠ C R I = m ∠ C A I .
m ∠ C R I = m ∠ R B I + m ∠ R I B . As R I = I A = R B , m ∠ R B I = m ∠ R I B , then m ∠ C R I = 2 m ∠ R B I = m ∠ R B A .
On the other hand, m ∠ C A I = m ∠ B A I , then m ∠ R B A = m ∠ I A B . △ A B I ≅ △ B A R ( S . A . S . ). Thus m ∠ C A B = 2 m ∠ A B C .
By going through the same process for a second time, we can get B I = I S = C S → △ B I S ≅ △ C S I → m ∠ A B C = 2 m ∠ A C B .
Therefore, m ∠ C A B : m ∠ A B C : m ∠ B C A = 4 : 2 : 1 , making m ∠ A B C = 2 × 7 1 8 0 ° = 7 3 6 0 ° .