in the given figure l is parallel to m . Show that ∠1 +∠2 -∠3 is 180°
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Given ,
l // m ,
produce BC intersects DE at F .
i ) l // m , BF is the transversal.
<1 + x = 180°
[sum of Interior angles same side of
the transversal are supplementary ]
x = 180° - <1 ----( 1 )
ii ) <2+ y = 180°
y = 180° - <2 ----( 2 )
[ Linear pair ]
iii ) In ∆CDF ,
<3 + x + y = 180° ----( 3 )
[ Angle sum property ]
Substitute ( 1 ) and ( 2 ) in ( 3 ) ,we get
<3 + 180° - <1 + 180° - <2 = 180°
=> 180° = 180° -180° + <1 + <2 - <3
=> 180° = <1 + <2 - <3
Hence , proved.
••••
l // m ,
produce BC intersects DE at F .
i ) l // m , BF is the transversal.
<1 + x = 180°
[sum of Interior angles same side of
the transversal are supplementary ]
x = 180° - <1 ----( 1 )
ii ) <2+ y = 180°
y = 180° - <2 ----( 2 )
[ Linear pair ]
iii ) In ∆CDF ,
<3 + x + y = 180° ----( 3 )
[ Angle sum property ]
Substitute ( 1 ) and ( 2 ) in ( 3 ) ,we get
<3 + 180° - <1 + 180° - <2 = 180°
=> 180° = 180° -180° + <1 + <2 - <3
=> 180° = <1 + <2 - <3
Hence , proved.
••••
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