In a triangle $A B C, A D$ is the median drawn on the side $B C$ is produced to $E$ such that $A D = E D$ prove that $A B E C$ is a parallelogram.

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Solution

Proof: AD is the median of $△ A B C$

Produce $A D$ to $E$ such that $A D = E D$
Join BE and CE.

Now In $Δ^{s} A B D$ and $E C D$
$B D = D C (D$ is the midpoints of $B C)$
$∠ A D B = ∠ E D C$ (vertically opposite angles)
$A D = E D$ (Given)

So $△ A B D ≅ △ E C D$ (SAS rule)

Therefore, $A B = C E (C P C T)$

also $∠ A B D = ∠ E C D$

These are interior alternate angles made by the transversal $\leftarrow \to B C$ with lines $\leftarrow \to A B$ and $\leftarrow \to C E$ .
$∴ \leftarrow \to A B ∥ \leftarrow \to C E$

Thus, in a Quadrilateral ABEC,

$A B ∥ C E$ and $A B = C E$

Hence ABEC is a parallelogram.

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ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such that $AB = BE$ and $AD = DF$ .

Prove that $△ BEC ≅ △ DCF$ .

In a triangle ABC, median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram. [2 MARKS]

△ A B C \sim △ P Q R, A (△

A B C) = 81 {c m}^{2}, A (△ P Q R) =

121 {c m}^{2}, B C = 6.3 c m

△ A B C \sim △ P Q R \dots \dots

\begin{matrix} ∴ \frac{A (△ A B C)}{A (△ P Q R)} = \frac{□}{{Q R}^{2}} \dots \dots \dots \dots ∴ \frac{□}{121} = \frac{(6.3)^{2}}{{Q R}^{2}} ∴ \frac{□}{11} = \frac{6.3}{Q R} \dots \dots . (Taking square root of both sides) Q R = \frac{6.3 \times 11}{□} Q R = □ c m \end{matrix}

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