Question

(a) In the given figure, PA, QB and RC are perpendicular to AC. If PA = x units, QB = y units and RC = z units, prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .

(b) Sides AB and BC and median AD of triangle ABC are respectively proportional to sides PQ and QR and median PM of $△ PQR$ . Show that $△ A BC \sim △ PQR$ .

Approach

Given $P A$ , $QB$ , $RC$ perpendicular to $A C$ with $P A = x$ , $QB = y$ , $RC = z$ . By similar triangles: $△ A BQ \sim △ A CR$ gives $\frac{A B}{A C} = \frac{QB}{RC} = \frac{y}{z}$ . Similarly $△ CBQ \sim △ C A P$ gives $\frac{BC}{A C} = \frac{QB}{P A} = \frac{y}{x}$ . Adding: $\frac{A B + BC}{A C} = y (\frac{1}{x} + \frac{1}{z}) = 1$ , so $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$ .

Step-by-step working

1
$△ A BQ \sim △ A CR \Rightarrow \frac{A B}{A C} = \frac{y}{z}$
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2
$△ CBQ \sim △ C A P \Rightarrow \frac{BC}{A C} = \frac{y}{x}$
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3
Add: $\frac{A B + BC}{A C} = \frac{y}{z} + \frac{y}{x} = y (\frac{1}{x} + \frac{1}{z})$
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4
Since $A B + BC = A C$ , we have $1 = y (\frac{1}{x} + \frac{1}{z})$
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5
Therefore $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$
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Concepts used

similar trianglesproveproportionalitymedianbasic proportionality theorem

Approach

Step-by-step working

Concepts used

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