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In the given figure, PA, QB and RC each ...

In the given figure, PA, QB and RC each is perpendicular to AC such that `PA=x, RC=y, QB=z, AB=a, and BC=b`
Prove that `(1)/(x)+(1)/(y)=(1)/(z)`

Text Solution

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`Pa bot AC and QB bot AC rArr QB||PA`.
Thus, in `Delta PAC, QB||PA`, So, `Delta QBC~ Delta PAC`.
`:. (QB)/(PA)=(BC)/(AC) rArr (z)/(x) =(b)/(a+b)....(i)` [ bu the propery of similar `Delta` ]
In `Delta RAC, QB|| RC`, So, `Delta QBA~ Delta RAC`.
`:. (QB)/(RC)=(AB)/(AC) rArr (z)/(y)=(a)/(a+b)...(ii)` [ by the propert of similar `Delta`]
From (i) and (ii), we get
`(z)/(x)+(z)/(y)=((b)/(a+b)+(a)/(a+b))=1`
`rArr (z)/(x)+(z)/(y)=1 rArr (1)/(x)+(1)/(x)+(1)/(y)=(1)/(z)`
Hence, `(1)/(x)+(1)/(y)=(1)/(z)`
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