Question

The sides AB and AC of triangle ABC are produced to points E and D respectively. if
bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove that
∠BOC = 90° -
1
2
∠BAC

Answer 1

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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Answer 2

Answer:

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

Step-by-step explanation: