In trapezium ABCD- AB in parallel to DC- P and Q the mid-points of AD and BC respectively- BP produced meets CD produced point E- Prove that-PQ is parallel to AB-

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Toppr Admin · Accepted Answer

For triangle -ECB- PQ-parallel CE-Again -CE-parallel AB-Therefore -PQ-parallel AB-Hence proved

In trapezium $$ABCD, AB$$ in parallel to $$DC; P$$ and $$Q$$ the mid-points of $$AD$$ and $$BC$$ respectively. $$BP$$ produced meets $$CD$$ produced at point $$E$$. Prove that:
$$PQ$$ is parallel to $$AB$$.

For triangle $$ECB:\ PQ\parallel CE$$
Again $$CE\parallel AB$$
Therefore $$PQ\parallel AB$$
Hence proved

In trapezium $$ABCD, AB$$ in parallel to $$DC; P$$ and $$Q$$ the mid-points of $$AD$$ and $$BC$$ respectively. $$BP$$ produced meets $$CD$$ produced at point $$E$$. Prove that:$$PQ$$ is parallel to $$AB$$.

For triangle $$ECB:\ PQ\parallel CE$$Again $$CE\parallel AB$$Therefore $$PQ\parallel AB$$Hence proved

In trapezium $$ABCD, AB$$ in parallel to $$DC; P$$ and $$Q$$ the mid-points of $$AD$$ and $$BC$$ respectively. $$BP$$ produced meets $$CD$$ produced at point $$E$$. Prove that:
$$PQ$$ is parallel to $$AB$$.

For triangle $$ECB:\ PQ\parallel CE$$
Again $$CE\parallel AB$$
Therefore $$PQ\parallel AB$$
Hence proved