Question

The sides AB and AC of triangle ABC are produced to points E and D respectively. If bisectors BO and CO of angle CBE and angle BCD respectively meet at point O, then prove that angle BOC = 90-half angle BAC​

Answer 1

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Answer 2

Solution: RAY BO IS THE BISECTOR OF angle CBE

therefore, angle CBO =1 by 2 of angle CBE

=1 by 2 (180degree - y)

= 90degree - y by 2 ................... (1)

similarly ray CO is the BISECTOR of angle BCD

therefore , angle BCO = 1 by 2 of angle BCD

= 1 by 2 (180degree -z)

=90degree - z by 2 ................. (2)

In triangle BOC ,

angle BOC + angle BCO + angle CBO =180DEGREE (angle sum property )...... (3)

substituting (1)and (2) In (3) we get

angle BOC +90DEGREE -z by 2+90degree -y by 2 =180DEGREE

angle BOC =z by 2 +y by 2

angle BOC =1 by 2 (y+z)............ (4)

x+y+z=180DEGREE ( angle sum property of a triangle )

y+z= 180DEGREE - x

therefore (4) becomes

angle BOC = 1 by 2 (180degree -x)

=90degree - x by 2

=90degree - 1 by 2 angle BAC

hence proved

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