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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 662.
∠ADC = ∠AEC ⇒ 180° - B = B + A/2 + C/2⇒ 180° - B = 90° + B/2 ⇒ 3B/2 = 90 °⇒ B = 60°⇒ ∠AFC = 120°
Since D is the orthocenter of tri. ABC so angle (ADC)=180- angle (B)Since E is the incenter of tri. ABC so angle (AEC)=B + (A+C)/2We have 180-B=B+(A+C)/2Replace (A+C)/2= 90- B/2 in above expression We get B= 60 and angle (AFC)=120Peter Tran