Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the tri.aTtlf le DEF are 90° -:jLA, 90° - jLB arµl 90° - 2 LC

L ADE = L ABE ...(i) [Angles in the same segment of a circle] L ADF = L ACF ...(ii) [Reason same as above} Now L EDF = L ADE + L ADF B C L EDF = L ABE + L ACF D ...(iii ) [From (i), (ii)] As BE and CF are bisectors of L ABC and L ACB. 1 1 :. L ABE = 2 L ABC and L ACF = 2 L ACB ...(iv) From (iii) and (iv), we get 1 1 1 L EDF = 2 L ABC + 2 L ACB = 2 (L ABC + L ACB) In triangle ABC, L ABC + L ACB + L BAC = 180° .•.(v) 166 MATHEMATICS- I X L ABC + L ACB = 180° - L BAC. Substituting in (v), we get LEDF = (180° - LBAC) = 90° - 2 LBAC = 90°- 2 LA Similarly, we can show that LDEF = 90° - 2 LB; LDFE = 90° - 2 LC