Question
In figure, show that AB || EF.
✓
Answer
Produce EF to intersect AC at K. Now,$\angle\text{DCE}+\angle\text{CEF}=35+145=180^\circ$
Therefore, EF || CD ... (1) [Since Sum of Co-interior angles is 180] Now,$\angle\text{BAC}=\angle\text{ACD}=57^\circ$
⇒ BA || EF ... (2) [Alternative angles are equal] From (1) and (2) AB || EF [Since, Lines parallel to the same line are parallel to each other] Hence proved.
Therefore, EF || CD ... (1) [Since Sum of Co-interior angles is 180] Now,$\angle\text{BAC}=\angle\text{ACD}=57^\circ$
⇒ BA || EF ... (2) [Alternative angles are equal] From (1) and (2) AB || EF [Since, Lines parallel to the same line are parallel to each other] Hence proved.
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