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Question
Bisectors of vertex angles A, B, and C of a triangle ABC intersect its circumcircle at the points D, E and F respectively. Prove that angle EDF = 90° – `1/2` ∠A.
Solution
We have,
Given that, BE is the bisector of B.
`therefore angle ABE = (angleB)/2`
`angleADE = angleABE` ...(Angles in the same segment for chord AF)
Similarly, `angle ACF = angleADF = (angleC)/2` ...(Angle in the same segment for chord AF)
`angleD = angleADE + angleADF`
`angleD = (angleB)/2 + (angleC)/2`
= `1/2 (angleB + angleC)`
= `1/2 (180^circ - angleA)` ...`(∴ angleA + angleB + angleC = 180^circ)`
`angleD = 90^circ - (angleA)/2`
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Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
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By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
