Question

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that 

(i) ∠A = ∠B 

(ii) ∠C = ∠D 

(iii) ∆ABC ≅ ∠∆BAD 

(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer 1

Let us extend \(AB\) and draw a line through \(C\), which is parallel to \(AD\), intersecting \(AE\) at point \(E\). It is clear that \(AECD\) is a parallelogram.

(i) Proving \( BC = CE \)

In parallelogram \(AECD\), opposite sides are equal, so:

\[ AD = CE \quad \text{(Opposite sides of parallelogram AECD)} \]

Also, we are given that \(AD = BC\). Therefore, we can deduce that:

\[ BC = CE \]

Next, we use the property of opposite angles in a parallelogram. Since \(AD = BC\), we know that:

\[ \angle CEB = \angle CBE \quad \text{(Angles opposite to equal sides are also equal)} \]

Since \(AD \parallel CE\) and \(AE\) is a transversal line, we have:

\[ \angle A + \angle CEB = 180^\circ \quad \text{(Angles on the same side of the transversal)} \]

Using the relation \( \angle CEB = \angle CBE \), we can rewrite this as:

\[ \angle A + \angle CBE = 180^\circ \quad \text{(Equation 1)} \]

Similarly, since \( \angle B + \angle CBE = 180^\circ \) (linear pair angles), we have:

\[ \angle B + \angle CBE = 180^\circ \quad \text{(Equation 2)} \]

From equations (1) and (2), we conclude:

\[ \angle A = \angle B \]

(ii) Proving \( AB \parallel CD \)

Now, consider the angles formed by the transversal lines. We know:

\[ \angle A + \angle D = 180^\circ \quad \text{(Angles on the same side of the transversal)} \]

Also, we have:

\[ \angle C + \angle B = 180^\circ \quad \text{(Angles on the same side of the transversal)} \]

Since \( \angle A = \angle B \) from the previous result, we can equate the two expressions:

\[ \angle A + \angle D = \angle C + \angle B \]

Therefore, \( \angle C = \angle D \), which proves that:

\[ AB \parallel CD \]

(iii) Proving \( \Delta ABC \cong \Delta BAD \)

In triangles \( \Delta ABC \) and \( \Delta BAD \), we know:

\[ AB = BA \quad \text{(Common side)} \]

\[ BC = AD \quad \text{(Given)} \]

\[ \angle B = \angle A \quad \text{(Proved before)} \]

Using the SAS congruence rule, we can conclude that:

\[ \Delta ABC \cong \Delta BAD \]

(iv) Conclusion

From the congruence of \( \Delta ABC \) and \( \Delta BAD \), we can deduce:

\[ AC = BD \quad \text{(By CPCT)} \]