**ABCD is a trapezium in which and AD=BC (see figure). Show that**

**(i)**

**(ii)**

**(iii)**

**(iv)** diagonal AC=diagonal BD

**Hint:** Extend AB and draw a line through C parallel to DA intersecting AB produced at E.

**Solution (i)**

ABCD is a trapezium in which we have AD=BC.

By construction, we extend AB and draw a line CE parallel to AD. **(1)**

We already have because ABCD is a trapezium.

** (2)**

From **(1)** and **(2)**, we can say that AECD is a parallelogram.

**(A quadrilateral is a parallelogram if both the pairs of opposite sides are parallel.)**

We have AD=BC ** (Given) ** ** (3)**

Also, we have AD=CE ** (Opposite sides of parallelogram are equal) (4)**

From **(3)** and **(4)**, we can say that BC=CE

** (In triangle, angles opposite to equal sides are equal) (5)**

We have **(Co-Interior angles, ) (6)**

And, **(Linear pair) (7)**

From **(6)** and** (7)**, we can say that

Using** (5)** in the above equation, we get

**Solution (ii)**

We have showed in **solution (i)** that **(8)**

But, we have **(Opposite angles in parallelogram are equal)** **(9)**

And, ** (Alternate Interior angles, )** ** (10)**

Using **(9)** and** (10)** in equation **(8)**, we get

**Solution (iii)**

Join AC and BD.

In and

BC=AD **(Given)**

** (Proved above)**

AB=BA ** (Common)**

Therefore, **by SAS congruence rule**, we have .

**Solution (iv)**

We showed in **solution (iii)** that

** (Corresponding parts of congruent triangles are equal)**