bisectors of the angle B and C of an isosceles triangle with AB equal to AC intersect each other at O . BO is produced to point M prove that angle MOC equal to angle ABC.
Answers
Answer:
Given:
Lines OB and OC are the bisectors of ∠B and ∠C of an isosceles ΔABC such that AB=AC which intersect each other at O and BO is produced to M.
To prove:
∠MOC=∠ABC
Consider the diagram shown below.
Proof:
In ΔABC,
AB=AC (given)
∠ACB=∠ABC (angles opposite to equal sides are equal)
½∠ACB=½∠ABC (dividing both sides by 2)
Therefore,
∠OCB=∠OBC …… (1)
(Since, OB and OC are the bisector of ∠B and ∠C)
Now, from equation (1), we have
∠MOC=∠OBC+∠OCB ∵ sum of interior angle = Exterior angle
∠MOC=2∠OBC
⇒∠MOC=2∠ABC× ½
=∠ABC
(Since, OB is the bisector of ∠B)
Hence, proved.
Answer:
∠MOC=∠ABC
Step-by-step explanation:
Given: Lines OB and OC are the bisectors of ∠B and ∠C of an isosceles ΔABC such that AB=AC which intersect each other at O and BO is produced to M.
To prove: ∠MOC=∠ABC
Consider the diagram shown below.
Proof: In ΔABC,
AB=AC (given)
∠ACB=∠ABC (angles opposite to equal sides are equal)
1/2∠ACB= 1/2∠ABC (dividing both sides by 2)
Therefore,
∠OCB=∠OBC …… (1) (Since, OB and OC are the bisector of ∠B and ∠C)
Now, from equation (1), we have
∠MOC=∠OBC+∠OCB ∵ sum of interior angle = Exterior angle
∠MOC=2∠OBC
⇒∠MOC=2∠ABC× 1/2=∠ABC (Since, OB is the bisector of ∠B)
Hence, proved.