In the given figure, $CD$ and $RS$ are respectively the medians of similar triangles $\Delta ABC$ and $\Delta PQR$. Prove that $\Delta ADC \sim \Delta PSR$.
Step-by-Step Solution
Key Concept: $\Delta ABC \sim \Delta PQR \Rightarrow \dfrac{AC}{PR} = \dfrac{AB}{PQ} = \dfrac{2AD}{2PS} = \dfrac{AD}{PS}$. Also $\angle A = \angle P$. By SAS, $\Delta ADC \sim \Delta PSR$.
$\Delta ABC \sim \Delta PQR \Rightarrow \angle A = \angle P$ and $\dfrac{AC}{PR} = \dfrac{AB}{PQ}$. [0.5 Mark]
Since $D, S$ are midpoints, $AB = 2AD, PQ = 2PS \Rightarrow \dfrac{AC}{PR} = \dfrac{2AD}{2PS} = \dfrac{AD}{PS}$. [1.0 Mark]
In $\Delta ADC$ and $\Delta PSR$, $\dfrac{AC}{PR} = \dfrac{AD}{PS}$ and $\angle A = \angle P$. By SAS, $\Delta ADC \sim \Delta PSR$. Proved! [0.5 Mark]
---
🎯 Official CBSE Marking Scheme:
Expressing $AB/PQ = AD/PS$: 1.0 Mark
Applying SAS similarity criterion: 1.0 Mark
Correct Answer: