In Figure AB-DE- AB-DE- AC-DF and AC-DF- Prove that BC-EF and BC-EF

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Theorem 5: One Pair of Opposite Sides Parallel and Equal

In Figure AB|...

Question

In Figure $ \mathrm{AB}\left|\right|\mathrm{DE}, \mathrm{AB}=\mathrm{DE}, \mathrm{AC}\left|\right|\mathrm{DF}$ and $ \mathrm{AC}=\mathrm{DF}.$ Prove that $ \mathrm{BC}\left|\right|\mathrm{EF}$ and $ \mathrm{BC}=\mathrm{EF}$

D A

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Solution

Step 1: Prove that $ABED$ is a parallelogram.
Given: $AB | | DE, AB = DE, AC | | DF$ and $AC = DF .$
$AB = DE and AB ∥ DE (Given)$
A quadrilateral having equal and parallel opposite sides is a parallelogram.
Therefore, $ABED$ is a parallelogram.
Step 2: Prove that $ADFC$ is a parallelogram.
$AC = DF and AC ∥ DF$ [ given ]
A quadrilateral having equal and parallel opposite sides is a parallelogram
Therefore, $ADFC$ is a parallelogram.
Step 3: Prove the required conditions.
We can write that
$AD = BE and AD ∥ BE - (1) (∴ ADEB is a parallelogram) AD = CF and AD ∥ CF - (2) (∴ ADFC is a parallelogram)$
From equations $(1)$ and $(2)$
$BE = CF and BE ∥ CF$
Hence proved that $BC | | EF$ and $BC = EF$ .

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In Figure $ \mathrm{AB}\left|\right|\mathrm{DE}, \mathrm{AB}=\mathrm{DE}, \mathrm{AC}\left|\right|\mathrm{DF}$ and $ \mathrm{AC}=\mathrm{DF}.$ Prove that $ \mathrm{BC}\left|\right|\mathrm{EF}$ and $ \mathrm{BC}=\mathrm{EF}$D A

In Figure $ \mathrm{AB}\left|\right|\mathrm{DE}, \mathrm{AB}=\mathrm{DE}, \mathrm{AC}\left|\right|\mathrm{DF}$ and $ \mathrm{AC}=\mathrm{DF}.$ Prove that $ \mathrm{BC}\left|\right|\mathrm{EF}$ and $ \mathrm{BC}=\mathrm{EF}$

D A