Question

Bisectors of the angles A,B and C of a triangle ABC intersect it's circumcircle at D,England and Friday respectively prove that the angles of a triangle DEF are 90°1/2A,90°1/2B and 90°1/2C

Answer 1

here is ur answer


It is given that BE is the bisector of ∠B

.: ∠ABE = ∠B/2

However, ∠ADE = ∠ABE (Angles in the same segment for chord AE)

∠ADE = ∠B/2

Similarly, ∠ACF = ∠ADF = ∠C/2 (Angle in the same segment for chord AF)

∠D = ∠ADE + ∠ADF

∠D = ∠B/2 + ∠C/2

= ½ ( ∠B + ∠C )

= ½ ( 180 – ∠A ) {since ∠A + ∠B + ∠C =180 ; angle sum property }
= 90 – ∠A/2

Similarly, it can be proved that

∠E = 90 – ∠B/2 &

∠F = 90 – ∠C/2


hope this helps u....

Answer 2

Answer:

Here, ABC is inscribed in a circle with center O and the bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.

Now, join DE, EF and FD

As angles in the same segment are equal, so,

∠FDA = ∠FCA ————-(i)

∠FDA = ∠EBA ————-(i)

Adding equations (i) and (ii) we have,

∠FDA + ∠EDA = ∠FCA + ∠EBA

Or, ∠FDE = ∠FCA + ∠EBA = (½)∠C + (½)∠B

We know, ∠A + ∠B + ∠C = 180°

So, ∠FDE = (½)[∠C + ∠B] = (½)[180° – ∠A]

⇒ ∠FDE = [90 – (∠A/2)]

In a similar way,

∠FED = [90 – (∠B/2)]

And,

∠EFD = [90 – (∠C/2)]