l and line segments AB, CD and EF are concurrent at point P. Prov…

Question

$\text{l}\parallel\text{m}$ and line segments AB, CD and EF are concurrent at point P. Proved that $\frac{\text{AE}}{\text{BF}}=\frac{\text{AC}}{\text{BD}}=\frac{\text{CE}}{\text{FD}}.$

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Answer

Given: $\text{l}\parallel\text{m}$
Line segments AB, CD and EF intersect at P.
Points A, E and C are on line l.
Points D, F and B are on line m.
To prove: $\frac{\text{AE}}{\text{BF}}=\frac{\text{AC}}{\text{BD}}=\frac{\text{CE}}{\text{FD}}\ .........(\text{i})$
Proof: In $\triangle\text{AEP}$ and $\triangle\text{BFP},$
$\text{l}\parallel\text{m}$ [Given]
$\angle1=\angle2$ [Alternate interior angles]
$\angle3=\angle4$ [Alternate interior angles]
$\therefore\triangle\text{AEP}\sim\triangle\text{BFP}$ [By AA simillarity criterion]
In $\triangle\text{CEP}$ and $\triangle\text{DFP},$
$\text{l}\parallel\text{m}$ [Given]
$\angle7=\angle8$ [Alternate interior angles]
$\angle5=\angle6$ [Alternate interior angles]
$\therefore\triangle\text{CEP}\sim\triangle\text{DFP}$ [By AA simillarity criterion]
$\Rightarrow\frac{\text{CE}}{\text{DF}}=\frac{\text{CP}}{\text{DP}}=\frac{\text{EP}}{\text{FP}}\ ........(\text{ii})$
In $\triangle\text{ACP}$ and $\triangle\text{BDP},$
$\text{l}\parallel\text{m}$ [Given]
$\angle1=\angle2$ [Alternate interior angles]
$\angle5=\angle6$ [Alternate interior angles]
$\therefore\triangle\text{ACP}\sim\triangle\text{BDP}$ [By AA simillarity criterion]
$\Rightarrow\frac{\text{AC}}{\text{BD}}=\frac{\text{AP}}{\text{BP}}=\frac{\text{CP}}{\text{DP}}\ ......(\text{iii})$
$\Rightarrow\frac{\text{AP}}{\text{PB}}=\frac{\text{AC}}{\text{BP}}=\frac{\text{CP}}{\text{DP}}=\frac{\text{CE}}{\text{DF}}=\frac{\text{EP}}{\text{FP}}=\frac{\text{AE}}{\text{BF}}$
$\Rightarrow\frac{\text{AC}}{\text{BD}}=\frac{\text{AE}}{\text{BF}}=\frac{\text{CF}}{\text{DF}}$
Hence, proved