For the following statments state whether true (T) or false(F):
The ratio of the areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
The ratio of the areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
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True.
Solution:
Given that $\triangle\text{ABC}\sim\triangle\text{DEF}$ in which AX and DY are the bisectors of $\angle\text{A}$ and $\angle\text{D}$ respectively.
To prove: $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AX}^2}{\text{DY}^2}$
We know that,
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the squares of the squares of the corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}\dots\text{(i)}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\angle\text{A}=\angle\text{D}$
$\Rightarrow\frac{1}{2}\angle\text{A}=\frac{1}{2}\angle\text{D}$
$\Rightarrow\angle\text{BAX}=\frac{1}{2}\angle\text{EDY}$
In $\triangle\text{ABX}$ and $\triangle\text{DEY},$
$\angle\text{BAX}=\angle\text{EDY}$ and $\angle\text{B}=\angle\text{E}$
So, by the AA criterion for similarity,
$\triangle\text{ABX}\sim\triangle\text{DEY}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{AX}}{\text{DY}}$
$\Rightarrow\frac{\text{AB}^2}{\text{DE}^2}=\frac{\text{AX}^2}{\text{DY}^2}\dots(\text{ii})$
From (i) and (ii),
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AX}^2}{\text{DY}^2}$
Solution:
Given that $\triangle\text{ABC}\sim\triangle\text{DEF}$ in which AX and DY are the bisectors of $\angle\text{A}$ and $\angle\text{D}$ respectively.
To prove: $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AX}^2}{\text{DY}^2}$
We know that,
The ratio of the areas of two similar triangles is equal to the ratio of the squares of the squares of the squares of the corresponding sides.
$\therefore\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AB}^2}{\text{DE}^2}\dots\text{(i)}$
$\triangle\text{ABC}\sim\triangle\text{DEF}$
$\Rightarrow\angle\text{A}=\angle\text{D}$
$\Rightarrow\frac{1}{2}\angle\text{A}=\frac{1}{2}\angle\text{D}$
$\Rightarrow\angle\text{BAX}=\frac{1}{2}\angle\text{EDY}$
In $\triangle\text{ABX}$ and $\triangle\text{DEY},$
$\angle\text{BAX}=\angle\text{EDY}$ and $\angle\text{B}=\angle\text{E}$
So, by the AA criterion for similarity,
$\triangle\text{ABX}\sim\triangle\text{DEY}$
$\Rightarrow\frac{\text{AB}}{\text{DE}}=\frac{\text{AX}}{\text{DY}}$
$\Rightarrow\frac{\text{AB}^2}{\text{DE}^2}=\frac{\text{AX}^2}{\text{DY}^2}\dots(\text{ii})$
From (i) and (ii),
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DEF})}=\frac{\text{AX}^2}{\text{DY}^2}$
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