the sides AB and AC of triangle ABC are produed to point 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 of angle BAC
Answers
- Let BO the Ray and also BO is the bisector of CBE
Hence, CBO = Half CBE(1/2)
Let CO the Ray and also the bisector of BCD
Hence, BCO = Half BCD(1/2
The Third equation:-
Adding all the triange we get 180°
BOC and BOC + BCO+ CBO=180° _equatiob - (3)
Adding substituting for equation 1,2 and 3
and ,
After adding all the angle we get 180°
Hence,equation (4)
Property used to solve this question:-
ASP (Angle Sum property)
Answer :
Proved.
Step-by-step explanation:
Given :
The sides AB and AC of triangle ABC are produed to point E and D.
Bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O.
To Prove :
Prove :
BO is the bisector of ∠CBE
So, ∠BCO = ∠DCO =
∠ CBE is the exterior angle of Δ ABC.
Implies,
⇒ ( ∠CBE ) = (x + y)
⇒ ∠ CBO = (x + y)
Same here,
∠BCD is the exterior angle of Δ ABC
So,
⇒ ∠ BCD = (a + b).
⇒ ( ∠BCD ) = (a + b)
⇒ ∠ BCD = (a + b)
In ΔOBC,
⇒ ∠ BOC + ∠ BCO + CBO = 180°
★__________{ Angle Sum Property }
⇒ ∠ BOC ( a + b) + (x + y) = 180°
⇒ ∠ BOC + ( a + b + x + y) = 180° .....Eq(1)
In Δ ABC
( a + b + x + y) = 180° .....Eq(2)
★__________{ Angle Sum Property. }
Putting values of Eq(2) in Eq(1),
⇒ ∠ BOC + (a + b + x + y) = 180°
⇒ ∠ BOC + ( a + 180°) = 180°
⇒ ∠ BOC + + × 180°
⇒ ∠ BOC = + 90° = 180°
⇒ ∠ BOC = 180° - 90° =
⇒ ∠ BOC = 90° -
⇒ ∠ BOC = 90° - × ∠ BAC
Hence, Proved!
Points to Remember -
- External angle is sum of two interior opposite angles.
- Angle Sum Property, i.e - x + y + z = 180°