Question

. In a triangle ABC, D is mid-point of BC; AD
is produced upto E so that DE = AD.
Prove that :
(i) A ABD and A ECD are congruent.
(ii) AB = EC.
(iii) AB is parallel to EC.​

Answer 1

$\huge{\longrightarrow{\rm{ANSWER}}{\longleftarrow}}$

Given :-

In a ∆ ABC , D is the mid-point of side BC ; AD is produced upto E such that DE = AD .

Required to prove :-

• ∆ ABD ≅ ∆ ECD

• AB = EC

• AB || EC

Construction :-

Join E to C and similarly join E to B

Concept used :-

• Congruency axioms

Proof :-

Given that :-

In a ∆ ABC , D is the mid point of side BC . AD is produced upto E such that DE = AD .

we need to prove that the ∆ ABD is congruent to ∆ ECD , AB = EC and

AB || EC

So,

Let's consider ∆ ABD and ∆ ECD

In ∆ ABD & ∆ ECD

AD = DE ( Side )

[ Reason :- Since it is given in the question ]

∠ADB = ∠EDC ( angle )

[ Reason :- Vertically opposite angles are equal ]

BD = CD ( side )

[ Reason :- Since D is the mid-point of side BC ]

Hence,

By using S.A.S. Congruency criteria

We can say that ,

∆ ABD ≅ ∆ ECD

Similarly,

We also say that ,

AB = EC

[ Reason :- C.P.C.T. ]

since,

AB = EC

So,

AB || EC

Points to remember :-

C.P.C.T. stands for Corresponding Parts of Congruent Triangles

Some of the Congruency axioms are

• AAA ( Angle, Angle, Angle property )
• SAS ( Side , Angle , Side )
• ASA ( Angle , side , Angle )
• RHS ( Right angle , Hypotenuse , side )
• SSS ( side , side , side )

Here, SAS rule is used only be consider when the include angle between the two sides is taken .

For example : Refer to the above answer

Otherwise it will considered as SSA rule of congruency .

Similarly, AAA congruency axiom is not mostly used in these proofs because atleast one side must be mentioned .

In the attachment the diagram is mentioned in which the triangles which were being considered were being shaded with pencil