O is a point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC.Through o,a line segment PG is drawn parallel to AB meeting AD in P and BC in Q.Prove that PO = QO
Hello anshulbshah,
If you draw your trapezium,
you will find that the areas of triangles BDC and ADC are equal, as the have the same bases and perpendicular heights.
Then use the fact that triangle APO is a reduced version of ADC
and triangle BOQ is a reduced version of BDC.
It follows that since their areas are equal and they have the same perpendicular heights,
then their bases PO and OQ must be equal.