In the figure given below- if BC - EF and FG - CD- AE-AB -

In the Δ ABC, EF ∥ BC ∴ AE/AB = AF/AC ......(1) [By BPT] In Δ ACD, FG ∥ CD ∴ AF/AC = AG/AD .......(2) [By BPT] From (1) and (2), we get AE/AB = AF/AC = ...

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Standard X

Mathematics

Basic Proportionality Theorem

In the figure...

Question

In the figure given below, if BC $∥$ EF and FG $∥$ CD. $\frac{A E}{A B}$ =

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Solution

In the $Δ$ ABC,
EF $∥$ BC
$∴$ $\frac{A E}{A B}$ = $\frac{A F}{A C}$ ......(1) [By BPT]

In $Δ$ ACD,
FG $∥$ CD
$∴$ $\frac{A F}{A C}$ = $\frac{A G}{A D}$ .......(2) [By BPT]

From (1) and (2), we get
$\frac{A E}{A B}$ = $\frac{A F}{A C}$ = $\frac{A G}{A D}$
$\Rightarrow \frac{A E}{A B}$ = $\frac{A G}{A D}$

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In the figure given below, if BC ∥ EF and FG ∥ CD. AEAB =

In the Δ ABC, EF ∥ BC ∴ AEAB = AFAC ......(1) [By BPT] In Δ ACD, FG ∥ CD ∴ AFAC = AGAD .......(2) [By BPT] From (1) and (2), we get AEAB = AFAC = AGAD ⇒AEAB = AGAD

In the figure given below, if BC $∥$ EF and FG $∥$ CD. $\frac{A E}{A B}$ =