Mathematics — Standard (041) — CBSE Class 10 board question

Given: PA ⊥ AC, QB ⊥ AC, RC ⊥ AC; AP = x, BQ = y, CR = z.

To Prove: $\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{y}$

Construction: Join PC and AR. Let them intersect BQ at points S and T respectively (as shown, Q lies on the intersection).

Proof:

Step 1: In △APC, BQ is perpendicular to AC and B lies on AC.

Since PA ∥ BQ (both ⊥ AC), by AA similarity criterion:

$$\triangle APB \sim \triangle QTB \quad \Rightarrow \quad \frac{BQ}{AP} = \frac{AB}{AC}$$

Wait — more directly, in △PAC, since QB ∥ PA (both ⊥ AC):

By Basic Proportionality / AA similarity: $\triangle ABQ \sim \triangle ACP$

$$\Rightarrow \quad \frac{BQ}{AP} = \frac{AB}{AC} \quad \Rightarrow \quad \frac{y}{x} = \frac{AB}{AC} \tag{1}$$

Step 2: In △RAC, since QB ∥ CR (both ⊥ AC):

By AA similarity: $\triangle QBC \sim \triangle RCA$ (actually △CBQ ~ △CAR)

$$\Rightarrow \quad \frac{BQ}{CR} = \frac{BC}{AC} \quad \Rightarrow \quad \frac{y}{z} = \frac{BC}{AC} \tag{2}$$

Step 3: Adding (1) and (2):

$$\frac{y}{x} + \frac{y}{z} = \frac{AB}{AC} + \frac{BC}{AC} = \frac{AB + BC}{AC} = \frac{AC}{AC} = 1$$

$$\Rightarrow \quad y\!\left(\frac{1}{x} + \frac{1}{z}\right) = 1$$

$$\boxed{\therefore \quad \frac{1}{x} + \frac{1}{z} = \frac{1}{y}}$$ Hence proved.

Source: Chapter 6 (Triangles), Sections 6.3 & 6.4 — AA Similarity Criterion

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• Key idea: PA ∥ BQ ∥ RC (all ⊥ to AC) lets you form two pairs of similar triangles using AA criterion (right angle + common/shared angle).
• From △PAC and △QBC, extract the ratios $\frac{y}{x} = \frac{AB}{AC}$ and $\frac{y}{z} = \frac{BC}{AC}$.
• Since AB + BC = AC, adding the two ratios gives 1, and dividing both sides by y gives the result.
• Examiners award marks for: stating the parallel lines, correctly identifying the two similar triangle pairs, writing the proportionality ratios, and the final addition step. Don't skip any step.