Question

1. If a line intersets "AB & AC of
A ABC at Dr E respectively &
parallel to BC prove that A - AC
AB AC​

Answer 1

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$\sf\underbrace{Correct\: Question: }$

• If a line intersects sides AB and AC of a ΔABC at D and E respectively and is parallel to BC, prove that $\sf\dfrac{AD}{AB}$=$\sf\dfrac{AE}{AC}$

$\sf\underline{Given: }$

• ΔABC, where line intersects sides AB and AC at D and E And DE || BC

$\sf\underline{To\:prove:}$

• $\sf\dfrac{AD}{AB}$=$\sf\dfrac{AE}{AC}$

$\sf\underline{Solution:}$

We know that if a line drawn parallel to one side of triangle, intersects the other two sides in distinct points, then it divides the other 2 side in same ratio.

$\sf\underline{Therefore :}$ $\sf\dfrac{AD}{DB}$=$\sf\dfrac{AE}{EC}$

ㅤㅤ$:⟹$ $\sf\dfrac{AD}{DB}$=$\sf\dfrac{AE}{EC}$

ㅤㅤ$:⟹$ $\sf\dfrac{DB}{AD}$=$\sf\dfrac{EC}{AE}$

$\sf\underline{Adding\:1\:on\: both\:Sides:}$

• ㅤㅤ$:⟹$$\sf\dfrac{DB}{AD}$+ 1 =$\sf\dfrac{EC}{AE}$ + 1

• ㅤㅤ$:⟹$$\sf\dfrac{DB\:+\:AD}{AD}$=$\sf\dfrac{EC\:+\:AE}{AE}$

• ㅤㅤ$:⟹$$\sf\dfrac{AB}{AD}$=$\sf\dfrac{AC}{AE}$

• ㅤㅤ$:⟹$$\sf\dfrac{AD}{AB}$=$\sf\dfrac{AE}{AC}$

$\underline{\purple{\bf \red{\dag}\:Hence\:Proved}}$

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Answer 2

Answer:

Correct Question

If a line intersects AB & AC of ΔABC at D and E respectively and DE || BC, prove that

$\bf \:\dfrac{AD}{AB} = \bf\dfrac{AE}{AC}$

Answer

Given:

• A ΔABC, where line intersects sides AB and AC at D and E and DE || BC

To prove:-

$\bf\dfrac{AD}{AB} =\bf\dfrac{AE}{AC}$

Solution:-

★In ΔABC and ΔADE

⇛/_ADE = /_ABC [Corresponding angles]

⇛/_AED = /_ACB [Corresponding angles]

∴ ΔABC ~ ΔADE [AA Similarity]

$\bf\implies \:\dfrac{AD}{AB} = \bf\dfrac{AE}{AC} \: (using \: CPST)$

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