In a triangle A B C , points X and Y are on A B and A C , respectively, such that X Y is parallel to

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In a $\triangle A B C$, points $X$ and $Y$ are on $A B$ and $A C$, respectively, such that $X Y$ is parallel to $B C$. Which of the two following equalities always hold? (Here $[P Q R]$ denotes the area of $\triangle P Q R)$.

$I$. $[B C X]=[B C Y]$

$II$. $[A C X] \cdot[A B Y]=[A X Y] \cdot[A B C]$

Neither $I$ nor $II$

Only $I$

Only $II$

Both $I$ and $II$

(KVPY-2015)

Solution

(d)

$A B C$ is a triangle points $X$ and $Y$ on $A B$ and $A C$ respectively.

$X Y$ is parallel to $B C$.

$I$. Area of $B C X:$ Area of $B C Y$

It is true because same base between same parallels.

$II$. Area of $\triangle A C X=\frac{1}{2}(A X)(A C) \sin A$

Area of $\triangle A B Y=\frac{1}{2}(A Y)(A B) \sin A$

$\therefore$ (Area of $\triangle A C X)$ (Area of $\triangle A B Y)$

$=\frac{1}{2}(A X)(A C) \sin A \times \frac{1}{2}(A Y)(A B) \sin A$

$=\frac{1}{2}(A X)(A Y) \sin A \times \frac{1}{2}(A B)(A C) \sin A$

$=(\text { Area of } \triangle A X Y) \text { (Area of } \triangle A B C)$

Hence,$I$ and $II$ both are true.

Standard 11

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normal

In a $\triangle A B C$, points $X$ and $Y$ are on $A B$ and $A C$, respectively, such that $X Y$ is parallel to $B C$. Which of the two following equalities always hold? (Here $[P Q R]$ denotes the area of $\triangle P Q R)$. $I$. $[B C X]=[B C Y]$ $II$. $[A C X] \cdot[A B Y]=[A X Y] \cdot[A B C]$

Solution

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