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Question
In the given figure, AB || CD. Prove that p +q -r =180.
In the given figure, AB || CD. Prove that p +q -r =180.
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Solution
Draw PFQ∥AB∥CD.
Now, PFQ∥AB and EF is the transversal.
Then,
∠AEF+∠EFP=180°.....(1)
Also, PFQ∥CD.
Angles on the same side of a transversal line are supplementary
∠PFG=∠FGD=r°(Alternate Angles)
and ∠EFP=∠EFG-∠PFG=q°-r°
putting the value of ∠EFP from eqn. (i)we get,
p°+q°-r°=180°
⇒p+q-r=180
Draw PFQ∥AB∥CD.
Now, PFQ∥AB and EF is the transversal.
Then,
∠AEF+∠EFP=180°.....(1)
Also, PFQ∥CD.
Angles on the same side of a transversal line are supplementary
∠PFG=∠FGD=r°(Alternate Angles)
and ∠EFP=∠EFG-∠PFG=q°-r°
putting the value of ∠EFP from eqn. (i)we get,
p°+q°-r°=180°
⇒p+q-r=180
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