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It is given figure, AB||DE|| and BD||EF....

It is given figure, `AB||DE|| and BD||EF`.
Prove that `DC^(2)=CFxxAC`.

Text Solution

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In `Delta ABC. AB=||DE`
`:. (CD)/(DA)=(CE)/(EB) " " ..(i)` [ by Thale's theorme]
In `Delta CDB,BD ||EF`
`:. (CF)/(FD)=(CE)/(EB) " ..(ii)` [ by Thale's theorme]
From (i) and (ii), we get
`(CD)/(DA)=(CF)/(FD)`
`rArr (DA)/(DC)=(FD)/(CF)" "` [ taking reciprocals]
`rArr (DA)/(DC)+1(FD)/(CF)+1`
`rArr (DA+DC)/(DC)=(FD+CF)/(CF)`
`rArr (AC)/(DC)=(DC)/(CF)`
`rArr DC^(2)=CFxxAC`
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