Question

Please answer this fast as possible

Answer 1

Step-by-step explanation:

$Given : PA\perp \: AC, \: QB\perp AC,\: \\RC\perp AC \\ To Prove: \: \frac{1}{x} + \frac{1}{z} = \frac{1}{y} \\ Proof: \\ in \: \triangle \: CQB \:\& \:\triangle\:CPA\\</p><p>\angle CBQ=\angle CAP.. (each \:90°)\\</p><p>\angle BCQ=\angle ACP.. (common \:angle)\\</p><p>\therefore \triangle \: CQB \:\sim \:\triangle\:CPA\\....(By\:AA\:test \:of\:similarity) \\</p><p>\therefore \frac {QB} {PA}= \frac {CB} {AC}... (csst)\\</p><p>\therefore \frac {y} {z}= \frac {CB} {AC}...(1)\\</p><p></p><p>\\ in \: \triangle \: AQB \:\& \:\triangle\:ARC\\</p><p>\angle ABQ=\angle ACR.. (each \:90°)\\</p><p>\angle BAQ=\angle CAR.. (common \:angle)\\</p><p>\therefore\triangle \: AQB \:\sim \triangle\:ARC \\....(By\:AA\:test \:of\:similarity) \\</p><p>\therefore \frac {QB} {RC}= \frac {AB} {AC}... (sast)\\</p><p>\therefore \frac {y} {x}= \frac {AB} {AC}...(2)$

Adding (1) & (2), we find :

$\frac {y} {z}+\frac {y} {x}= \frac {CB} {AC}+ \frac {AB} {AC}\\\therefore</p><p>y(\frac {1} {z}+\frac {1} {x})=\frac {AB+CB} {AC}\\\therefore y(\frac {1} {z}+\frac {1} {x})=\frac {AC} {AC}\\\therefore y(\frac {1} {z}+\frac {1} {x})=1</p><p>\\\therefore \frac {1} {x}+\frac {1} {z}=\frac {1} {y}$

Thus Proved