Hypotenuse $AC$ of a right triangle $ABC$ is divided at $D$ into two segments of length $a$ and $b$. Perpendiculars $DX$ and $DY$ are drawn from $D$ to $AB$ and $BC$ respectively. Prove that:
(i) $DX \cdot BC = a \cdot AB$ (or similar ratio relation)
(ii) $DX^2 + DY^2 = DE^2$ (or prove $\Delta ADX \sim \Delta DYC$).
Step-by-Step Solution
Key Concept: In right $\Delta ABC$ with $DX \perp AB$ and $DY \perp BC$, $DXBY$ is a rectangle. $\Delta ADX \sim \Delta DYC \sim \Delta ABC$.
Since $DX \perp AB$, $DY \perp BC$, and $\angle B = 90^\circ$, quadrilateral $DXBY$ is a rectangle. Thus $BX = DY$ and $BY = DX$. [1.0 Mark]
In $\Delta ADX$ and $\Delta DYC$:
1. $\angle AXD = \angle DYC = 90^\circ$.
2. $\angle ADX = \angle C$ (corresponding angles as $DX \parallel BC$). [1.5 Marks]
By AA similarity, $\Delta ADX \sim \Delta DYC$. [1.0 Mark]
Therefore $\dfrac{AX}{DY} = \dfrac{DX}{YC} = \dfrac{AD}{DC} = \dfrac{a}{b}$. [1.0 Mark]
Cross-multiplying gives $b \cdot AX = a \cdot DY$ and $b \cdot DX = a \cdot YC$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Identifying $DXBY$ as a rectangle: 1.0 Mark
Establishing angle equalities $\angle AXD = \angle DYC = 90^\circ$ and $\angle ADX = \angle C$: 1.5 Marks
Proving similarity $\Delta ADX \sim \Delta DYC$: 1.0 Mark
Writing side ratios and cross-multiplying: 1.5 Marks
Correct Answer: