If a line intersects sides AB and AC of a triangle ABC at D and E respectively and is parallel to BC, prove that, AB/DB=AC/EC.
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Answer:
\frac{AD}{AB}=\frac{AE}{AC}
AB
AD
=
AC
AE
Step-by-step explanation:
Concept used:
Basic proportionality theorem(Thales theorem):
If a line is drawn parallel to one side of a triangle then it cuts other two sides proportionally.
In ΔABC, DE || BC
By Thales theorem,
\frac{AD}{DB}=\frac{AE}{EC}
DB
AD
=
EC
AE
Taking reciprocals
\frac{DB}{AD}=\frac{EC}{AE}
AD
DB
=
AE
EC
Adding 1 on both sides
\begin{lgathered}1+\frac{DB}{AD}=1+\frac{EC}{AE}\\\\\frac{AD+DB}{AD}=\frac{AE+EC}{AE}\\\\\frac{AB}{AD}=\frac{AC}{AE}\end{lgathered}
1+
AD
DB
=1+
AE
EC
AD
AD+DB
=
AE
AE+EC
AD
AB
=
AE
AC
Taking reciprocals once again, we get
\frac{AD}{AB}=\frac{AE}{AC}
AB
AD
=
AC
AE
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