# In the given figure, AB||CD. Prove that p+q-r=180

Given AB||CD, AEF = P°, EFG = q°, FGD =

Draw a line FH||AB||CD

HFG = FGD = r° [Because HF||CD and alternate angles] ___________ (i)

EFH = EFG - HFG

EFH = q – r ______________________ (i)

AEF + EFH = 180° [Because AB||HF]

AEF + EFH = 180°

p + (q – r) = 180°

p + q – r = 180°Proved.

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