# Ex.9.3 Q13 Areas of Parallelograms and Triangles Solution - NCERT Maths Class 9

## Question

\(ABCD\) is a trapezium with \(AB\; || \;DC\). A line parallel to \(AC\) intersects \(AB\) at \(X\) and \(BC\) at \(Y\).

Prove that \(ar (ADX) = ar (ACY)\).

[Hint: Join \(CX\).]

## Text Solution

**What is known?**

\(ABCD\) is a trapezium with \(AB || DC\). A line parallel to \(AC\) intersects \(AB\) at \(X\) and \(BC\) at \(Y.\)

**What is unknown?**

How we can prove that \(ar (ADX) = ar (ACY).\)

**Reasoning:**

We can use theorem for triangles \(ADX\) and \(ACX\), if two triangles are on same base and between same pair of parallel lines then both will have equal area. Similarly for triangles we can use for triangles \(ACX\) and \(ACY\). Now by comparing both the results, we will get the required result.

**Steps:**

It can be observed that \(\begin{align}\Delta ADX \, \rm {and} \,\Delta ACX \end{align}\) lie on the same base \(AX\) and are between the same parallels \(AB\) and \(DC\).

According to Theorem 9.2 : **Two triangles on the same base (or equal bases) and between the same parallels are equal in area.**

\[\begin{align}\therefore \text {ar }(\Delta {ADX})\!=\!\text {ar}(\Delta {ACX})\dots{(1)}\end{align}\]

\(\Delta ACY\) and \(\Delta ACX\) lie on the same base \(AC\) and are between the same parallels \(AC\) and \(XY\).

According to Theorem 9.2: **Two triangles on the same base (or equal bases) and between the same parallels are equal in area.**

\[\begin{align}(\Delta {ACY})\!=\!\text {ar}({ACX})\,\ldots \text (2)\end{align}\]

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

\[\begin{align}{\rm{ar}}\left( {{{\Delta ADX}}} \right)\!=\!{\rm{ar}}\left( {{{\Delta ACY}}} \right)\end{align}\]