Question

Seg AB and seg DC are perpendicular to seg bc. seg ac and seg bd intersect each other at P. seg PQ perpendicular to seg BC. AB=x, PQ=y, DC=z. Prove that:1/x+1/z=1/y​

Answer 1

$\text{In \triangle{BQP} }$

$tan\theta_1=\frac{y}{BQ}$......(1)

$\text{In \triangle{BCD} }$

$tan\theta_1=\frac{z}{BC}$......(2)

$\text{In \triangle{PQC} }$

$tan\theta_2=\frac{y}{QC}$......(3)

$\text{In \triangle{ABC} }$

$tan\theta_2=\frac{x}{BC}$......(4)

$\text{Now,}$

$BC=BQ+QC$

$\text{Using (1) and (3)}$

$BC=\frac{y}{tan\theta_1}+\frac{y}{tan\theta_2}$

$\text{Using (2) and (4)}$

$BC=\frac{y}{z/BC}+\frac{y}{x/BC}$

$BC=\frac{y\;BC}{z}+\frac{y\;BC}{x}$

$\implies\;1=\frac{y}{z}+\frac{y}{x}$

$\implies\;1=y[\frac{1}{z}+\frac{1}{x}]$

$\implies\boxed{\bf\frac{1}{y}=\frac{1}{x}+\frac{1}{z}}$