In a trapezium \(ABCD\) with \(AB\parallel DC\), diagonals \(AC\) and \(BD\) meet at \(O\). Through \(O\), draw \(PQ\parallel AB\) meeting \(AD\) at \(P\) and \(BC\) at \(Q\). Prove that \(PO=QO\).
Proved.
Step 1: We are given a trapezium \(ABCD\) in which \(AB \parallel DC\). The diagonals \(AC\) and \(BD\) meet at \(O\). A line \(PQ\) is drawn through \(O\) parallel to \(AB\) and \(DC\). This line cuts \(AD\) at \(P\) and \(BC\) at \(Q\).
Step 2: Since \(PQ \parallel AB \parallel DC\), the angles made by these parallel lines with the diagonals will be equal. So, in triangles \(\triangle AOP\) and \(\triangle COQ\): - \(\angle PAO = \angle QCO\) (alternate interior angles) - \(\angle APO = \angle QCO\) (corresponding angles) Thus, \(\triangle AOP \sim \triangle COQ\) (by AA similarity).
Step 3: From similarity, we get the ratio of corresponding sides: \[ \dfrac{PO}{QO} = \dfrac{AO}{CO} \quad ...(1) \]
Step 4: Similarly, consider triangles \(\triangle BOQ\) and \(\triangle DOP\). Since \(PQ \parallel AB \parallel DC\), these two triangles are also similar (AA similarity). Therefore, \[ \dfrac{QO}{PO} = \dfrac{BO}{DO} \quad ...(2) \]
Step 5: A very important property of a trapezium: The diagonals of a trapezium intersect each other proportionally. This means, \[ \dfrac{AO}{CO} = \dfrac{BO}{DO} \quad ...(3) \]
Step 6: From (1), we have \(\dfrac{PO}{QO} = \dfrac{AO}{CO}\). From (2), we have \(\dfrac{QO}{PO} = \dfrac{BO}{DO}\). But from (3), \(\dfrac{AO}{CO} = \dfrac{BO}{DO}\). So, \[ \dfrac{PO}{QO} = \dfrac{QO}{PO} \]
Step 7: Cross multiplying gives: \[ PO^2 = QO^2 \] Taking square roots, we get: \[ PO = QO \]
Final Answer: Hence, it is proved that \(PO = QO\).