Question
$ABCD$ is a trapezium with $AB || DC. E$ and $F$ are two points on non-parallel sides $AD$ and $BC$ respectively, such that $EF$ is parallel to $AB$. Show that $\frac{AE}{ED}=\frac{BF}{FC}$
✓
Answer
Given, In trapezium $ABCD,$
$AB || DC$ and $EF || DC$
To prove $\frac{AE}{ED}=\frac{BF}{FC}$
Construction: Join $AC$ to intersect $EF$ at $G.$
Proof Since, $AB || DC$ and $EF || DC$
$EF || AB$ [since, lines parallel to the same line are also parallel to each other ]$...... (i)$
In $\triangle ADC$, $EG || DC$ [$\because $ $EF || DC$]
By using basic proportionality theorem,
$\frac{AE}{ED}=\frac{AG}{GC} ....(ii)$
In $\triangle ABC$, $ GF || AB$ [$\because $ $EF || AB$ from $(i)]$
By using basic proportionality theorem ,
$\frac{CG}{AG}=\frac{CF}{BF}$ or $\frac{AG}{GC}=\frac{BF}{CF}$ [ On taking reciprocal of the terms]$............. (iii)$
From Equations $(ii)$ and $(iii)$, we get
$\frac{AE}{ED}=\frac{BF}{FC}$
Hence Proved.
$AB || DC$ and $EF || DC$
To prove $\frac{AE}{ED}=\frac{BF}{FC}$
Construction: Join $AC$ to intersect $EF$ at $G.$
Proof Since, $AB || DC$ and $EF || DC$
$EF || AB$ [since, lines parallel to the same line are also parallel to each other ]$...... (i)$
In $\triangle ADC$, $EG || DC$ [$\because $ $EF || DC$]
By using basic proportionality theorem,
$\frac{AE}{ED}=\frac{AG}{GC} ....(ii)$
In $\triangle ABC$, $ GF || AB$ [$\because $ $EF || AB$ from $(i)]$
By using basic proportionality theorem ,
$\frac{CG}{AG}=\frac{CF}{BF}$ or $\frac{AG}{GC}=\frac{BF}{CF}$ [ On taking reciprocal of the terms]$............. (iii)$
From Equations $(ii)$ and $(iii)$, we get
$\frac{AE}{ED}=\frac{BF}{FC}$
Hence Proved.
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