In a trapezium $ABCD$, $AB \parallel DC$ and its diagonals intersect at $O$. Show that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.
Step-by-Step Solution
Key Concept: Draw line through $O$ parallel to $AB$, use BPT in $\Delta ABC$ and $\Delta ABD$. Alternatively, prove $\Delta AOB \sim \Delta COD$.
Stepwise Solution:
In $\Delta AOB$ and $\Delta COD$:
$\angle OAB = \angle OCD$ (Alternate interior angles as $AB \parallel DC$)
$\angle OBA = \angle ODC$ (Alternate interior angles as $AB \parallel DC$)
$\angle AOB = \angle COD$ (Vertically opposite angles). [1.0 Mark]
By AAA similarity criterion, $\Delta AOB \sim \Delta COD$. [1.0 Mark]
Therefore $\dfrac{AO}{CO} = \dfrac{BO}{DO} \Rightarrow \dfrac{AO}{BO} = \dfrac{CO}{DO}$. Proved! [1.0 Mark]
Marking Scheme:
• Identifying angle equalities: 1.0 Mark
• Establishing similarity $\Delta AOB \sim \Delta COD$: 1.0 Mark
• Rearranging side ratios to get $AO/BO = CO/DO$: 1.0 Mark
Correct Answer: