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Solution of Triangle

In a triangle...

Question

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 following two 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)

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(I) only

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(II) only

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Both (I) and (II)

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Solution

The correct option is D Both (I) and (II)

$[B C X] = [B C Y]$ as both the triangles have same base and same height.

(II) $ar (△ A C X) = \frac{1}{2} A C . A X sin A$
$ar (△ A B Y) = \frac{1}{2} A B . A Y sin A$
$ar (△ A X Y) = \frac{1}{2} A X . A Y sin A$
$ar (△ A B C) = \frac{1}{2} A B . A C sin A$
$\Rightarrow ar (△ A C X) \cdot ar (△ A B Y) = ar (△ A X Y) \cdot ar (△ A B C)$
$i.e., [A C X] \cdot [A B Y] = [A X Y] \cdot [A B C]$

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[B C X] = [B C Y]

(I I)

[A C X] . [A B Y] = [A X Y] . [A B C]

$△$ ABC is an acute angled triangle. DE is drawn parallel to BC as shown. Which of the following are always true?

i) $△ A B C \sim △$ ADE

ii) $\frac{A D}{B D} = \frac{A E}{E C}$

iii) $D E = \frac{B C}{2}$

In triangle ABC; AB = AC. P, Q and R are mid-points of sides AB, AC and BC respectively.

Prove that :
(i) PR = QR (ii) BQ = CP

D and E are points on the sides AB and AC respectively of a $△$ ABC such that DE || BC. Which of the following statement is true?

(i) $△$ ADE ~ $△$ ABC

(ii) $\frac{a r e a o f △ A D E}{a r e a o f △ A B C}$ = $\frac{A D^{2}}{A B^{2}}$

(iii) $\frac{a r e a o f △ A D E}{a r e a o f △ A B C}$ = $\frac{A B^{2}}{A C^{2}}$

Δ A B Y ≅ Δ A C X

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