Question

the sides AB & AC of a triangle are produced to the points E & D respectively . BO & CO are bisectors of angle CBE & BCD INTERSECTING at O. prove that angle CBE = 90 DEGREE -1/2 ANGLE BAC.

Answer 1

 OB = OC

(ii) AO bisects A.

Ans. (i) ABC is an isosceles triangle in which AB = AC.

 C = B [Angles opposite to equal sides]

 OCA + OCB = OBA + OBC

 OB bisects B and OC bisects C

 OBA = OBC and OCA = OCB

 OCB + OCB = OBC + OBC

 2OCB = 2OBC

 OCB = OBC

Now in OBC,

OCB = OBC [Prove above]

 OB = OC [Sides opposite to equal sides]

(ii) In AOB and AOC,

AB = AC [Given]

OBA = OCA [Given]

And B = C

 B = C

 OBA = OCA

 OB = OC [Prove above]

 AOB AOC [By SAS congruency]

 OAB = OAC [By C.P.C.T.]

Hence AO bisects A.