Diagonals of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Through $O$, a line segment $PQ$ is drawn parallel to $AB$ meeting $AD$ in $P$ and $BC$ in $Q$. Prove that $PO = OQ$.
Step-by-Step Solution
Key Concept: In $\Delta ABD$, $PO \parallel AB \Rightarrow \dfrac{PO}{AB} = \dfrac{DP}{DA}$. In $\Delta ABC$, $OQ \parallel AB \Rightarrow \dfrac{OQ}{AB} = \dfrac{CQ}{CB}$. Show $\dfrac{DP}{DA} = \dfrac{CQ}{CB}$.
Stepwise Solution:
In $\Delta ABD$, $PO \parallel AB \Rightarrow \Delta DPO \sim \Delta DAB \Rightarrow \dfrac{PO}{AB} = \dfrac{DP}{DA}$ -- (1). [1.0 Mark]
In $\Delta ABC$, $OQ \parallel AB \Rightarrow \Delta CQO \sim \Delta CBA \Rightarrow \dfrac{OQ}{AB} = \dfrac{CQ}{CB}$ -- (2). [1.0 Mark]
In trapezium $ABCD$ with $POQ \parallel AB \parallel DC$, by BPT on non-parallel sides $AD$ and $BC$: $\dfrac{DP}{DA} = \dfrac{CQ}{CB}$.
From (1) and (2): $\dfrac{PO}{AB} = \dfrac{OQ}{AB} \Rightarrow PO = OQ$. Proved! [1.0 Mark]
Marking Scheme:
• Similarity in $\Delta DPO \sim \Delta DAB \Rightarrow PO/AB = DP/DA$: 1.0 Mark
• Similarity in $\Delta CQO \sim \Delta CBA \Rightarrow OQ/AB = CQ/CB$: 1.0 Mark
• Equating ratios via BPT to conclude $PO = OQ$: 1.0 Mark
Correct Answer: