Question

In $ \mathrm{\Delta ABC} $and$ \mathrm{\Delta DEF}$, $ \mathrm{AB} = \mathrm{DE}, \mathrm{AB} \left|\right| \mathrm{DE}, \mathrm{BC} = \mathrm{EF}$ and $ \mathrm{BC} \left|\right| \mathrm{EF}.$ Vertices $ \mathrm{A}$, $ \mathrm{B}$, and $ \mathrm{C}$ are joined to vertices $ \mathrm{D}$, $ \mathrm{E}$, and $ \mathrm{F}$ respectively (see Fig.). Show that (i) quadrilateral $ \mathrm{ABED}$ is a parallelogram (ii) quadrilateral $ \mathrm{BEFC}$ is a parallelogram (iii) $ \mathrm{AD} \left|\right| \mathrm{CF} \mathrm{and} \mathrm{AD} = \mathrm{CF}$ (iv) quadrilateral $ \mathrm{ACFD}$ is a parallelogram (v)$ \mathrm{AC} = \mathrm{DF}$ (vi) $ \mathrm{\Delta ABC} \cong \mathrm{\Delta DEF}.$

Answer 1

Step 1: Prove $\mathrm{ABED}$ is a parallelogram.

Given: $\mathrm{AB}=\mathrm{DE},\mathrm{AB}\left|\right|\mathrm{DE},\mathrm{BC}=\mathrm{EF}\mathrm{and}\mathrm{BC}\parallel \mathrm{EF}$

Since, $\mathrm{AB}=\mathrm{DE}\mathrm{and}\mathrm{AB}\parallel \mathrm{DE}$

So, opposite sides of the quadrilateral $\mathrm{ABED}$ are equal and parallel.

Therefore $\mathrm{ABED}$ is a parallelogram.

Step 2: Prove $\mathrm{BEFC}$ is a parallelogram

Since, $\mathrm{BC}=\mathrm{EF}\mathrm{and}\mathrm{BC}\parallel \mathrm{EF}$

So, opposites sides of the quadrilateral $\mathrm{BEFC}$ are equal and parallel.

Therefore $\mathrm{BEFC}$ is a parallelogram.

Step 3: Prove $\mathrm{AD}\left|\right|\mathrm{CF}\mathrm{and}\mathrm{AD}=\mathrm{CF}$

Since $\mathrm{ABED}$ is a parallelogram (Proved above)

$\therefore \mathrm{AD}=\mathrm{BE}\mathrm{and}\mathrm{AD}\parallel \mathrm{BE}-\left(1\right)$

Since $\mathrm{BEFC}$ is a parallelogram (Proved above)

$\therefore \mathrm{BE}=\mathrm{CF}\mathrm{and}\mathrm{BE}\parallel \mathrm{CF}-\left(2\right)$

From (1) and (2)

$\therefore \mathrm{AD}=\mathrm{CF}\mathrm{and}\mathrm{AD}\parallel \mathrm{CF}$

Step 4: Prove $\mathrm{ACFD}$ is a parallelogram

Since, $\mathrm{AD}=\mathrm{CF}\mathrm{and}\mathrm{AD}\parallel \mathrm{CF}$

So, opposite sides of the quadrilateral $\mathrm{ACFD}$ are equal and parallel.

Therefore $\mathrm{ACFD}$ is a parallelogram.

Step 5: Prove $\mathbf{}\mathrm{AC}=\mathrm{DF}$

Since $\mathrm{ACFD}$ is a parallelogram (Proved above).

So, opposite sides of the parallelogram are equal and parallel.

$\therefore \mathrm{AC}=\mathrm{DF}$

Step 6: Prove $\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.$

$$\begin{gathered} \mathrm{In}∆\mathrm{ABC}\mathrm{and}∆\mathrm{DEF} \\ \mathrm{AB}=\mathrm{DE}\left(\mathrm{Given}\right) \\ \mathrm{BC}=\mathrm{EF}\left(\mathrm{Given}\right) \\ \mathrm{AC}=\mathrm{DF}(\mathrm{Proved}\mathrm{in}\mathrm{Step}5) \end{gathered}$$

$\therefore ∆\mathrm{ABC}\cong ∆\mathrm{DEF}$ [ By SSS congruence rule ]

Hence proved that

(i) quadrilateral $\mathrm{ABED}$ is a parallelogram

(ii) quadrilateral $\mathrm{BEFC}$ is a parallelogram

(iii) $\mathrm{AD}\left|\right|\mathrm{CF}\mathrm{and}\mathrm{AD}=\mathrm{CF}$

(iv) quadrilateral $\mathrm{ACFD}$ is a parallelogram

(v)$\mathrm{AC}=\mathrm{DF}$

(vi) $\mathrm{\Delta ABC}\cong \mathrm{\Delta DEF}.$