Question 14 Marks
Read the following text carefully and answer the questions that follow:
Haresh and Deep were trying to prove a theorem. For this they did the following
$i.$ Draw a triangle $ABC$
$ii. \ D$ and E are found as the mid points of $AB$ and $AC$
$iii.\ DE$ was joined and $DE$ was extended to $F$ so $DE = EF$
$iv. \ FC$ was joined.
Questions :
$i. \triangle ADE$ and $\triangle EFC$ are congruent by which criteria?
$ii.$ Show that $C F \| A B$.
$iii.$ Show that $C F=B D$.
OR
Show that $D F=B C$ and $D F \| B C$.
Haresh and Deep were trying to prove a theorem. For this they did the following
$i.$ Draw a triangle $ABC$
$ii. \ D$ and E are found as the mid points of $AB$ and $AC$
$iii.\ DE$ was joined and $DE$ was extended to $F$ so $DE = EF$
$iv. \ FC$ was joined.
Questions :
$i. \triangle ADE$ and $\triangle EFC$ are congruent by which criteria?
$ii.$ Show that $C F \| A B$.
$iii.$ Show that $C F=B D$.
OR
Show that $D F=B C$ and $D F \| B C$.
Answer
View full question & answer→$\text { i. } \triangle ADE \text { and } \triangle CFE$
$DE = EF \text { (By construction) }$
$\angle AED =\angle CEF \text { (Vertically opposite angles) }$
$AE = EC ( By \text { construction) }$
$By \text { SAS criteria } \triangle ADE \cong \triangle CFE $
$ii. \triangle ADE \cong \triangle CFE$
Corresponding part of congruent triangle are equal
$\angle EFC=\angle EDA$
alternate interior angles are equal
$\Rightarrow AD \| FC$
$\Rightarrow CF \| AB $
$iii. \triangle ADE \cong \triangle CFE$
Corresponding part of congruent triangle are equal.
$CF=AD$
We know that $D$ is mid point $A B$
$\Rightarrow AD = BD$
$\Rightarrow CF = BD $
OR
$DE =\frac{B C}{2}($line drawn from mid points of $2$ sides of $\triangle$ is parallel and half of third side$)$
$DE \| BC$ and $DF \| BC$
$DF = DE + EF$
$\Rightarrow DF =2 DE ( BE = EF )$
$\Rightarrow DF = BC $
$DE = EF \text { (By construction) }$
$\angle AED =\angle CEF \text { (Vertically opposite angles) }$
$AE = EC ( By \text { construction) }$
$By \text { SAS criteria } \triangle ADE \cong \triangle CFE $
$ii. \triangle ADE \cong \triangle CFE$
Corresponding part of congruent triangle are equal
$\angle EFC=\angle EDA$
alternate interior angles are equal
$\Rightarrow AD \| FC$
$\Rightarrow CF \| AB $
$iii. \triangle ADE \cong \triangle CFE$
Corresponding part of congruent triangle are equal.
$CF=AD$
We know that $D$ is mid point $A B$
$\Rightarrow AD = BD$
$\Rightarrow CF = BD $
OR
$DE =\frac{B C}{2}($line drawn from mid points of $2$ sides of $\triangle$ is parallel and half of third side$)$
$DE \| BC$ and $DF \| BC$
$DF = DE + EF$
$\Rightarrow DF =2 DE ( BE = EF )$
$\Rightarrow DF = BC $
