Question

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

Answer 1

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:

Step-by-step explanation:

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