Question
In $\triangle\text{ABC}$ (Fig.), if $\angle1=\angle2,$ prove that $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}.$
✓
Answer
Given: A $\triangle\text{ABC}$ in which $\angle1=\angle2$To prove: $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
Construction: Draw CE || DA to meet BA produced in E.
Proof: since, CE || DA and AC cuts them.
$\therefore\angle2=\angle3\ ....(\text{i})$ [Alternate angles]
And, $\angle1=\angle4\ ...(\text{ii})$ [Corresponding angles]
But, $\angle1=\angle2$ [Given]
Form (i) and (ii), we get
$\angle3=\angle4$
Thus, in $\triangle\text{ACE},$ we have
$\angle3=\angle4$
$\Rightarrow\text{AE}=\text{AC}\ ...(\text{iii})$ [Sides opposite to equal angles are equal]
Now, In $\triangle\text{BCE},$ we have
DA || CE
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{BA}}{\text{AE}}$ [Using basic proportionality theorem]
$\Rightarrow\frac{\text{BD}}{\text{DC}}=\frac{\text{AB}}{\text{AC}}$ [$\because$ BA - AB and AE - AC from (iii)]
Hence, $\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}}$
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