Question
In Fig. the sides $BA$ and $CA$ have been produced such that: $BA = AD$ and $CA = AE$. Prove that segment $DE \| BC.$
✓
Answer
Given that, the sides $BA$ and $CA$ have been produced such that $BA = AD$ and $CA = AE$ and given to prove $DE \| BC$ Consider triangle $BAC$ and $DAE,$
We have $BA = AD and CA = AE$ [given in the data] And
also $\angle\text{BAC}=\angle\text{DAE}$ [vertically opposite angles]
So, by $SAS$ congruence criterion,
we have $\angle\text{BAC}\simeq\angle\text{DAE}$
$\text{BC}=\text{DE}$ and $\angle\text{DEA}=\angle\text{BCA},\angle\text{EDA}=\angle\text{CBA}$ [Corresponding parts of congruent triangles are equal]
Now, $DE$ and $BC$ are two lines intersected by a transversal that
$\angle\text{DEA}=\angle\text{BCA}\text{ i}.\text{e}..$
Alternate angles are equal Therefore, $DE, BC \| BC.$
We have $BA = AD and CA = AE$ [given in the data] And
also $\angle\text{BAC}=\angle\text{DAE}$ [vertically opposite angles]
So, by $SAS$ congruence criterion,
we have $\angle\text{BAC}\simeq\angle\text{DAE}$
$\text{BC}=\text{DE}$ and $\angle\text{DEA}=\angle\text{BCA},\angle\text{EDA}=\angle\text{CBA}$ [Corresponding parts of congruent triangles are equal]
Now, $DE$ and $BC$ are two lines intersected by a transversal that
$\angle\text{DEA}=\angle\text{BCA}\text{ i}.\text{e}..$
Alternate angles are equal Therefore, $DE, BC \| BC.$
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