If BC - EF and FG - CD then- AEAB- -

In ΔABC, EF || BC ∴AEAB=AFAC.... (1) (Basic proportinality theorem) In ΔACD, FG || CD ∴AFAG=AGAD.... (2) (Basic proportionality theorem) From 1 amp; 2, AEAB= ...

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Standard VII

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Rani Abbakkadevi

If BC || EF a...

Question

If BC || EF and FG || CD then, $\frac{A E}{A B} =$ ?

\frac{C F}{A F}

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\frac{G D}{A G}

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\frac{A G}{G D}

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\frac{A G}{A D}

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Solution

The correct option is D $\frac{A G}{A D}$
In $Δ$ ABC,
EF || BC
$∴ \frac{A E}{A B} = \frac{A F}{A C}$ .... (1)
(Basic proportinality theorem)
In $Δ$ ACD,
FG || CD
$∴ \frac{A F}{A G} = \frac{A G}{A D}$ .... (2)
(Basic proportionality theorem)
From 1 & 2,
$\frac{A E}{A B} = \frac{A F}{A C} = \frac{A G}{A D}$

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If BC || EF and FG || CD then, AEAB= ?

The correct option is D AGADIn ΔABC, EF || BC ∴AEAB=AFAC.... (1) (Basic proportinality theorem) In ΔACD, FG || CD ∴AFAG=AGAD.... (2) (Basic proportionality theorem) From 1 & 2, AEAB=AFAC=AGAD

If BC || EF and FG || CD then, $\frac{A E}{A B} =$ ?