Question

In the figure X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg `PQ||` set DE, set `QR||` set EF. Fill in the blanks to prove that set `PR||` seg DF.

Answer 1

In `DeltaXDE,PQ||DE`
`:.(XP)/(square)=(square)/(QE)=……………………..(square)`
………….(I) (Basic proportionality theorem)
In `DeltaXEF,QR||EF……(square)`
`:.(square)/(square)=(square)/(square)`
….(II) `(square)`
`:.(square)/(square)=(square)/(square)`....[From I and II]
`:.` set `PR||` set DF
gt......(Converset of basic proportionality theorem)
In `DeltaXDE, PQ||DE`...........Given
`:.(XP)/(PD)+(XQ)/(QE)`
......I (Basic proportionality theorem)
In `DeltaXEF,QR||E` Given
`:.(XQ)/(QE)=(XR)/(RF)`
...II (Basic proportionality theorem)
`:.(XP)/(PD)=(XR)/(RF)`.........From I and II
`:.` set `PR||` seg DF
..(Covnerset of basic proportionality theorem)