Answer :

**To Prove:** BF = BC

**Given:** AD is produced to E, DE = DC

EC is produced to meet AB in F.

**Concept Used:**

**Opposite sides of a parallelogram are equal.**

**ASA Congruence:** The **ASA rule** states that. If two angles and the included side of one triangle are equal to two angles and included a side of another triangle, then the triangles are **congruent**.

**Diagram:**

**Explanation:**

In Δ *ACE*, *D,* and *O* are mid points of *AE* and *AC* respectively.

∴ *DO*||*EC*

⇒ *OB*||*CF*

⇒ *AB* = *BF* ….(i)

⇒ *DC* = *BF* [*AB* = *DC* as *ABCD* is a parallelogram]

In Δ’s *EDC* and *CBF*, we have

⇒ *DC* = *BC*

⇒ *EDC* = *CBF*

and *ECD* = *CFB*

So, by ASA congruence criterion, we have

ΔEDC ≅ ΔCBF

⇒ *DE* = *BC*

⇒ *DC* = *BC*

⇒ *AB* = *BC*

⇒ *BF* = *BC* [*AB* = *BF* from (i)]

**Hence, Proved.**

Rate this question :

The figure formedRS Aggarwal & V Aggarwal - Mathematics

A <imRS Aggarwal & V Aggarwal - Mathematics