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Geometry Level 5

Let O O be the incenter of A B C \triangle ABC .

If B C = A C + A O BC = AC+AO and B C = 1 6 \angle B-\angle C=16^\circ , find A \angle A .


The answer is 98.

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1 solution

Choi Chakfung
May 13, 2016

P r o d u c e C A t o Z w h e r e A Z = A O L e t A b e θ . A O Z = A Z O = 180 ° ( 180 ° θ 2 ) 2 = θ 4 O C = O C , O C A = O C B , C Z = C B C Z O C B O C Z O = C B O = θ 4 ( c o r r . o f ) B = θ 2 , B C = 16 ° C = θ 2 16 ° A + B + C = 180 ° θ + θ 2 16 ° + θ 2 = 180 θ = 98 ° Produce\quad CA\quad to\quad Z\quad where\quad AZ=AO\\ Let\quad \angle A\quad be\quad \theta \quad .\angle AOZ=\angle AZO=\frac { 180°-(180°-\frac { \theta }{ 2 } ) }{ 2 } =\frac { \theta }{ 4 } \\ OC=OC,\angle OCA=\angle OCB,CZ=CB\Rightarrow \triangle CZO\cong CBO\\ CZO=CBO=\frac { \theta }{ 4 } (corr.\angle \quad of\cong \triangle )\\ \angle B=\frac { \theta }{ 2 } ,\angle B-\angle C=16°\Rightarrow \angle C=\frac { \theta }{ 2 } -16°\\ \angle A+\angle B+\angle C=180°\\ \theta +\frac { \theta }{ 2 } -16°+\frac { \theta }{ 2 } =180\\ \theta =98°

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