# Ex.8.1 Q12 Quadrilaterals Solution - NCERT Maths Class 9

## Question

\(ABCD\) is a trapezium in which \(AB\; ||\; CD\) and \(AD = BC\) (see the given figure). Show that

(i) \(\begin{align} &{ \angle {A}=\angle {B}} \end{align}\)

(ii) \(\begin{align} & { \angle {C}=\angle {D}} \end{align}\)

(iii) \(\begin{align}&{ \triangle {ABC} \cong \triangle {BAD}} \end{align}\)

(iv) diagonal \(AC =\) diagonal \(BD\)

[*Hint*: Extend \(AB\) and draw a line through \(C\) parallel to \(DA\) intersecting \(AB\) produced at \(E\).]

## Text Solution

**What is known?**

\(ABCD\) is a trapezium in which \(AB || CD\) and \(AD = BC.\)

**What is unknown?**

How we can show that

(i) \(∠A = ∠B\)

(ii) \(∠C = ∠D\)

(iii) \(∆ABC ≅ ∆BAD\)

(iv) diagonal \(AC =\) diagonal \(BD\)

**Reasoning:**

Consider parallel lines \(AD\) and \(CE, AE\) is the transversal line for them then sum of co-interior angles will be \(180\) degree also angles \(CBE\) and \(CBA\) are linear pairs gives sum \(180\), angles \(CEB\) equals to \(CBE\) because opposite to equal sides in triangle \(BCE\). Using these observations angle A is equal to angle B. Then AB is parallel to CD will help to show angle C is equal to angle D. Also, by using suitable congruence criterion we can show triangles congruent then we can say corresponding parts of congruent triangles will be equal.

**Steps:**

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

(i) \(AD = CE\) (Opposite sides of parallelogram \(AECD\))

However, \({AD}={BC}(\text { Given })\)

Therefore, \({BC}={CE}\)

\[\begin{align}&\angle {{CEB}} = \angle {{CBE}}\\&\left( \begin{array}{l}{\text{Angle opposite to equal }}\\{\text{sides are also equal}}\end{array} \right)\end{align}\]

Consider parallel lines and . is the transversal line for them.

\[\begin{align}&\angle {{A}} + \angle {{CEB}} = {180^\circ }\\&\left( \begin{array}{l}{\text{Angles on the same }}\\{\text{side of transversal}}\end{array} \right)\end{align}\]

\[\begin{align}&\angle {{A}} + \angle {{CBE}} = {180^\circ }\\&\left( \begin{array}{l}{\text{Using the relation }}\\\angle {{CEB}} = \angle {{CBE}}\end{array} \right) \ldots (1)\end{align}\]

However, \(\angle B\) + \(\angle CBE\) = (Linear pair angles) ... (2)

From Equations (1) and (2), we obtain

\(\angle A\) = \(\angle B\)

(ii)

\[\begin{align}&\angle A + \angle D = {180^\circ }\\&\left( \begin{array}{l}{\text{Angles on the same side}}\\{\text{ of the transversal}}\end{array} \right)\end{align}\]

Also,

\[\begin{align}&\angle C + \angle B = {180^\circ }\\&\left( \begin{array}{l}{\text{Angles on the same side }}\\{\text{of the transversal}}\end{array}\right)\end{align}\]

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

However,

\[\begin{align}&\angle A = \angle B\\&\left[ \begin{array}{l}{\text{Using the result }}\\{\text{obtained in (i) }}\end{array} \right]\\&\therefore \,\,\,\angle C = \angle D\end{align}\]

(iii) In \(\begin{align}\triangle {ABC} \text { and } \triangle {BAD}\end{align}\)

\[\begin{align}{{AB}} &= {{BA \text{(Common side) }}}\\{{BC}}& = AD\text {(Given) }\\\angle {{B}} &= \angle A {\text{(Proved before) }}\\\therefore \Delta {{ABC}} &\cong \Delta {{BAD }}\\&\left( {{\text{SAS congruence rule}}} \right)\end{align}\]

(iv) We had observed that,

\[\begin{align} \Delta {ABC}& \cong \Delta {BAD} \\ \;\;\;\;\therefore {AC}&= {{BD}({By} \;{CPCT})}\end{align}\]