In figure, $CD$ and $RS$ are respectively the medians of $\Delta ABC$ and $\Delta PQR$. If $\Delta ABC \sim \Delta PQR$, prove that:
(i) $\Delta ADC \sim \Delta PSR$
(ii) $\dfrac{CD}{RS} = \dfrac{AB}{PQ}$
Step-by-Step Solution
Key Concept: Use $\Delta ABC \sim \Delta PQR \Rightarrow \dfrac{AC}{PR} = \dfrac{AB}{PQ} = \dfrac{2 AD}{2 PS} = \dfrac{AD}{PS}$ and $\angle A = \angle P$.
Stepwise Solution:
(i) Given $\Delta ABC \sim \Delta PQR \Rightarrow \angle A = \angle P$ and $\dfrac{AC}{PR} = \dfrac{AB}{PQ}$. [0.5 Mark]
Since $CD, RS$ are medians, $D, S$ are midpoints of $AB, PQ \Rightarrow AB = 2AD$ and $PQ = 2PS$. [1.0 Mark]
Thus $\dfrac{AC}{PR} = \dfrac{2AD}{2PS} = \dfrac{AD}{PS}$. In $\Delta ADC$ and $\Delta PSR$: $\angle A = \angle P$ and $\dfrac{AC}{PR} = \dfrac{AD}{PS}$.
By SAS similarity criterion, $\Delta ADC \sim \Delta PSR$. [1.0 Mark]
(ii) From $\Delta ADC \sim \Delta PSR \Rightarrow \dfrac{CD}{RS} = \dfrac{AD}{PS} = \dfrac{AB}{PQ}$. Proved! [0.5 Mark]
Marking Scheme:
• Using median properties $AB=2AD, PQ=2PS$: 1.0 Mark
• Proving SAS similarity for $\Delta ADC \sim \Delta PSR$: 1.0 Mark
• Deducing median ratio equal to side ratio: 1.0 Mark
Correct Answer: