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Answer» the side of BC of TRIANGLE ABC is produced to D If the bisector of angle A meets BC in E. prove that angle abc + angle acd = 2 angle AEC Solution:the bisector of angle A meets BC in E => ∠BAE = ∠CAE = x Let SAY ∠ABC = ∠ABE = α ( as E is on BC) ∠AEC = ∠ABE + ∠BAE => ∠AEC = α + x ∠ACD = ∠AEC + ∠CAE => ∠ACD = α + x + x ∠ABC + ∠ACD = α + α + x + x => ∠ABC + ∠ACD = 2 ( α + x) => ∠ABC + ∠ACD = 2∠AEC |
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