prove that the bisector of two adjacent supplementary angles include a right angle
Answers
given
⠀⠀⠀⠀CE−→−CE→ is the bisector of ∠ACD and
⠀⠀⠀⠀⠀⠀⠀⠀⠀CF−→−CF→ is the bisector of ∠BCD
⠀⠀⠀To prove÷∠ECF = 90o
proof÷
From the figure we know that
∠ACD and ∠BCD form a linear pair of angles
⠀⠀⠀⠀⠀⠀⠀⠀⠀So we can write it as ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀∠ACD + ∠BCD = 180o
We can also write it as
⠀⠀⠀⠀⠀⠀∠ACE + ∠ECD + ∠DCF + ∠FCB = 180o
⠀⠀⠀⠀⠀⠀⠀⠀From the figure we also know that
∠ACE = ∠ECD and ∠DCF = ∠FCB
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀So it can be written as
⠀⠀⠀⠀⠀⠀∠ECD + ∠ECD + ∠DCF + ∠DCF = 180o
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
On further calculation we get
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀2 ∠ECD + 2 ∠DCF = 180o
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
Taking out 2 as common we get
⠀⠀⠀⠀⠀⠀2 (∠ECD + ∠DCF) = 180o
⠀⠀⠀⠀⠀⠀⠀By division we get
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀(∠ECD + ∠DCF) = 180/2
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀∠ECD + ∠DCF = 90o
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀Therefore, it is proved that ∠ECF = 90o
Answer:
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