AD is the median of triangle ABC. The ratio of area of ABD : area of ACD is _______.

2 : 1

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1 : 2

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1 : 3

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1 : 1

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If ABCD is a prallelogram, then prove the $a r e a (△ A B D) = a r e a (△ B C D) = a r e a (△ A B C) = a r e a (△ A C D) = \frac{1}{2} a r e a (| |^{g m} A B C D)$

Q. In a triangle

A B C, E

is the mid-point of median

A D

. Show that

a r e a (B E D) = \frac{1}{4} a r e a (A B C)

AD is the median of △ ABC and DE is the median of △ ABD. If area of △ BED : area of △ ABC =_____.

If CD is a median of $Δ$ ABC, then find the ratio of Area $(Δ A D C)$ to Area $(Δ C D B)$ .

A D

is a median of

△ A B C

and

P

is a point on

A C

such that

Area (△ A D P) : Area (△ A B D) = 2 : 3

and

Area (△ P D C) : Area (△ A B C) = 1 : a

. Then the value of

a

is:

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