Question

In a triangle ABC, D is mid-point of BC; AD is produced upto E so that DE = AD. Prove that :

(i) $\Delta$ ABD and $\Delta$ ECD are congruent.

(ii) AB = EC.

(iii) AB is parallel to EC.

Answer 1

Ok First of all,
According to Question, we find the figure as above.
Now in $\Delta$ ABD and $\Delta$ECD..
1. BD = CD........... as D is the midpoint of BC,
2. AD = ED.......... as given in Question,
3. Angle ADB = Angle EDC ...... Vertically Opposite Angles.....
So bye SAS Criteria of Congruency...
$\Delta$ABD ≈ $\Delta$ ECD.....

Now Second,
to prove; AB is parallel to EC...
for that,
in Question, D is the midpoint of AE as well as BC.
Or BC bisects AE or vice versa so it is clear that the figure form by joining ABEC is a Rhombus, Parallelogram Square or a Rectangle...
And in all those shapes the opposite sides are parallel...
Hence AB is parallel to EC