In a quadrilateral, CO and DO are the bisector of angle C and angle D respectively. Prove that angle A+angle B=2angle COD
Here x = 1/2 D
And y = 1/2C
Answers
Answered by
3
To proof:angleA+AngleB =2 angleCOD
angleA+AngleB+angleC+angleD=360°
1/2angleA+1/2AngleB+1/2angleC+1/2angleD=180°........(eq.1)
In ∆COD
1/2AngleC+1/2angleD+angleCOD=180°.........(eq.2)
Equate eq. 1 and 2
1/2angleA+1/2angleB+1/2angleC+1/2angleD=1/2angleC+1/2angleD+AngleCOD
1/2(AngleA+angleB)=AngleCOD
angleA+angleB=2angleCOD
Hence, Proved
angleA+AngleB+angleC+angleD=360°
1/2angleA+1/2AngleB+1/2angleC+1/2angleD=180°........(eq.1)
In ∆COD
1/2AngleC+1/2angleD+angleCOD=180°.........(eq.2)
Equate eq. 1 and 2
1/2angleA+1/2angleB+1/2angleC+1/2angleD=1/2angleC+1/2angleD+AngleCOD
1/2(AngleA+angleB)=AngleCOD
angleA+angleB=2angleCOD
Hence, Proved
Similar questions
1/2(<A+ <B) + x + y = 180°.........1
x + y +<COD = 180°..............2
By 1 and 2
1/2(<A + <B) +x +y = x + y + <COD
1/2(<A+<B) = <COD