Question

In the given figure, $AB\parallel PQ\parallel CD$, $AB=x$ units, $CD=y$ units and $PQ=z$ units, prove that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$.

Answer 1

In $\Delta ABD$ and $\Delta PQD$

$\angle$$D$$=$ $\angle$$D$ (common)

$\angle BAD=\angle QPD$ (corresponding angles)

So, $\Delta ABD\sim \Delta PQD$ (by AA similarity criterion)

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QD}$ (Ratio of corresponding sides of two similar triangles)

$\Rightarrow \frac{x}{z}=\frac{BQ+QD}{QD}$

$\Rightarrow \frac{x}{z}=\frac{BQ}{QD}+1$

$\frac{x}{z}-1=\frac{BQ}{QD}$

$\Rightarrow \frac{x-z}{z}=\frac{BQ}{QD}$ ---(i)

Similarly in $\Delta CBD$ and $\Delta PBQ$

$\angle B=\angle B$ (common)

$\angle BCD=\angle BPQ$ (corresponding angles)

So,$\Delta CBD\sim \Delta PBQ$ (by AA similarity criterion)

$\Rightarrow \frac{CD}{PQ}=\frac{BD}{BQ}$ (Ratio of corresponding sides of two similar triangles)

$\Rightarrow \frac{y}{z}=\frac{BQ+QD}{BQ}$

$\Rightarrow \frac{y}{z}=1+\frac{QD}{BQ}$

$\Rightarrow \frac{y-z}{z}=\frac{QD}{BQ}$

$\Rightarrow \frac{z}{y-z}=\frac{BQ}{QD}$ --(ii)

from (i) and (ii)

$\frac{x-z}{z}=\frac{z}{y-z}$

$\Rightarrow \left(x-z\right)\left(y-z\right)=z^2$

$\Rightarrow xy-yz-xz+z^2=z^2$

$xy=xz+yz$

$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z}$