In $\triangle ABC$, if $\angle 1=\angle 2$, prove that $\frac{AB}{AC}=\frac{BD}{DC}$.
"
Given:
In $\triangle ABC$, $\angle 1=\angle 2$.
To do:
We have to prove that $\frac{AB}{AC}=\frac{BD}{DC}$.
Solution:
Construction: Draw $CE \parallel DA$ to meet $BA$ produced at $E$.
$CE \parallel DA$ and $AC$ is the transversal.
Therefore,
$\angle 2=\angle 3$.......(i) (Alternate angles)
$\angle 1=\angle 4$.......(ii) (Corresponding angles)
$\angle 1=\angle 2$.......(iii)
From (i), (ii) and (ii),
$\angle 3=\angle 4$
This implies,
In $\triangle ACE$,
$AE=EC$ (Angles opposite to equal sides are equal)
In $\triangle BCE$,
$DA \parallel CE$
$\Rightarrow \frac{BD}{DC}=\frac{BA}{AE}$ (By basic proportionality theorem)
$\Rightarrow \frac{BD}{DC}=\frac{BA}{AC}$ (Since $AE=EC$)
$\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}$
Hence proved.
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