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

Given: PA, QB, RC ⊥ AC with PA = x, QB = y, RC = z. B lies between A and C on line AC.

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

Construction: Join PR. Let PR intersect QB at D.

Proof:

In △PAC and △DBC:
∠PAC = ∠DBC = 90° (given perpendiculars)
∠PCA = ∠DCB (common angle at C)

∴ △PAC ~ △DBC (AA similarity criterion)

$$\therefore \frac{DB}{PA} = \frac{BC}{AC} \implies \frac{DB}{x} = \frac{BC}{AC} \quad \cdots(1)$$

In △RCA and △DAB:
∠RCA = ∠DAB = 90°
∠RAC = ∠DAB ... ∠ACR = ∠ABD (common angle at A ... wait)

∠RCA = ∠DBA = 90°; ∠RAC = ∠DAB (common angle A)

∴ △RCA ~ △DBA (AA similarity criterion)

$$\therefore \frac{DB}{RC} = \frac{AB}{AC} \implies \frac{DB}{z} = \frac{AB}{AC} \quad \cdots(2)$$

Adding (1) and (2):

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

$$\therefore DB\left(\frac{1}{x} + \frac{1}{z}\right) = 1 \quad \cdots(3)$$

In △QBP and △QBR (or using △APR and △QBsame approach):

Since QB || PA || RC (all perpendicular to AC), QB = y and D lies on QB:

By similar triangles △PAC ~ △DBC and △RCA ~ △DBA, we showed D is on QB.
Since QB ⊥ AC and D is on QB, DB = QB = y.

Substituting in (3):

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

$$\boxed{\dfrac{1}{x} + \dfrac{1}{z} = \dfrac{1}{y}} \quad \textbf{Hence proved.}$$

Source: Chapter 6, Section 6.4 – Criteria for Similarity of Triangles

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• The key idea is to join PR (diagonal) and let it meet QB (extended to line through B perpendicular to AC) at point D, so DB = QB = y.
• Two pairs of AA-similar triangles are formed: △PAC ~ △DBC gives DB/x = BC/AC, and △RCA ~ △DBA gives DB/z = AB/AC.
• Adding these two ratios uses AB + BC = AC, giving DB(1/x + 1/z) = 1, and since DB = y the result follows.
• Examiners award marks for: correct construction (1), each similarity with reason (1+1), the addition step (1), and conclusion (1). Always state the AA criterion explicitly.