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Equal Angles Subtend Equal Sides

In the figure...

Question

In the figure sides $A B, B C, C A$ of $△ A B C$ are produced upto points $R, F, P$ respectively such that $A B = B R, B C = C P, C A = A F$
Prove that : $A (△ P E R) = 7 A (△ A B C)$

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Solution

Given that Sides $A B, B C a n d C A$ of $△ A B C$ are produced up to points $R, P, F$ respectively such that $A B = B R, B C = C P, a n d C A = A F$

Construction: Join $P A, F B a n d R C$

Let $A r e a (△ A B C) = a$

We know that median of a triangle divides it into triangles of equal area.

In $△ P A B, A C$ is the median
$\Rightarrow A r e a (△ A B C) = A r e a (△ A P C) = a$

In $△ P C F, P A$ is the median

$\Rightarrow A r e a (△ A P C) = A r e a (△ P F A) = a$
In $△ B C F, B A$ is the median

$\Rightarrow A r e a (△ A B C) = A r e a (△ A B F) = a$

In $△ R A F, F B$ is the median

$\Rightarrow A r e a (△ A B F) = A r e a (△ R B F) = a$

In $△ A C R, B C$ is the median

$\Rightarrow A r e a (△ A B C) = A r e a (△ R B C) = a$
In $△ P R B, R C$ is the median

$\Rightarrow A r e a (△ R B C) = A r e a (△ R P C) = a$

Thus, $A r e a (△ A B C) = A r e a (△ A P C) = A r e a (△ P F A) = A r e a (△ A B F) = A r e a (△ R B F) = A r e a (△ R B C) = A r e a (△ R P C)$
Hence $A r e a (△ P R F) = a + a + a + a + a + a + a = 7 a = 7 (A r e a △ A B C)$

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Similar questions

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In the figure sides AB,BC,CA of △ABC are produced upto points R,F,P respectively such that AB=BR,BC=CP,CA=AFProve that : A(△PER)=7A(△ABC)

In the figure sides $A B, B C, C A$ of $△ A B C$ are produced upto points $R, F, P$ respectively such that $A B = B R, B C = C P, C A = A F$
Prove that : $A (△ P E R) = 7 A (△ A B C)$