# In Fig. 8.11, AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.

**Solution:**

Given, AB || DE and AB = DE

AC || DF and AC = DF

We have to prove that BC || EF and BC = EF.

Considering __quadrilateral__ ABED,

Given, AB || DE

AB = DE

So, ABED is a __parallelogram__

Now, AD || BE

AD = BE ------------------------- (1)

Considering quadrilateral ACFD,

AC || FD

AC = FD

So, ACFD is a parallelogram

Now, AD || FC

AD = FC -------------------------- (2)

From (1) and (2),

AD = BE = FC

CF || BE

So, BCFE is a parallelogram.

We know that the opposite sides of a parallelogram are parallel and equal.

Now, BC = EF

BC || EF

Therefore, it is proven that BC = EF and BC || EF.

**✦ Try This: **In the given fig. AB ∣∣ CD, ∠BDC = 40° and ∠BAD = 75. Find x,y,z.

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 8

**NCERT Exemplar Class 9 Maths Exercise 8.4 Problem 9**

## In Fig. 8.11, AB || DE, AB = DE, AC || DF and AC = DF. Prove that BC || EF and BC = EF.

**Summary:**

A parallelogram is a quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure. In Fig. 8.11, AB || DE, AB = DE, AC || DF and AC = DF. It is proven that BC || EF and BC = EF

**☛ Related Questions:**

- E is the mid-point of a median AD of ∆ABC and BE is produced to meet AC at F. Show that AF = 1/3 AC
- Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is . . . .
- E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Pro . . . .

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