From the given diagram, in which ABCD is a parallelogram, ABL is … - Vidyadip

Question

From the given diagram, in which $\text{ABCD}$ is a parallelogram, $\text{ABL}$ is a line segment and $\mathrm{E}$ is mid$-$point of $\mathrm{BC}$.Prove that:$(i) \triangle \mathrm{DCE} \cong \triangle \mathrm{LBE};(ii)\mathrm{AB}=\mathrm{BL};(iii)\mathrm{AL}=2 \mathrm{DC}$

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Answer

Given: $\text{ABCD}$ is a parallelogram in which is the mid$-$point of $B C$. We need to prove that
$(i)\triangle \mathrm{DCE} \cong \triangle \mathrm{LBE}$
$(ii) \mathrm{AB}=\mathrm{BL}$.
$(iii) \mathrm{AL}=2 \mathrm{DC}$

$(i)$ In $\triangle \mathrm{DCE}$ and $\triangle \mathrm{LBE}$
$\angle \mathrm{DCE}=\angle \mathrm{EBL} \ldots[\mathrm{DC} \| \mathrm{AB}$, alternate angels $]$
$\mathrm{CE}=\mathrm{EB} \ldots[\mathrm{E}$ is the midpoint of $\mathrm{BC}]$
$\angle \mathrm{DEC}=\angle \mathrm{LEB} \quad\dots...[$ vertically opposite angels$]$
$\therefore$ By Angel$-$SIde$-$Angel Criterion of congruence, we have,
$\triangle \mathrm{DCE} \cong \triangle \mathrm{LBE}$
The corresponding parts of the congruent triangles are congruent.
$\therefore \mathrm{DC}=\mathrm{LB} \ldots[ [\text{c. p. c.t} ]\ldots .(1)$
$(ii) \mathrm{DC}=\mathrm{AB} \ldots[$ opposite sides of a parallelogram$]...(2)$
From $(1)$ and $(2), Ab = BL \ldots(3)$
$(iii) \mathrm{Al}=\mathrm{AB}+\mathrm{BL}\ldots (4)$
From $(3)$ and $(4), A l=A B+A B$
$\Rightarrow A L=2 A B$
$\Rightarrow \mathrm{AL}=2 \mathrm{DC}\dots..[$ From $(2)]$

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