Suppose-that-the-point-M-lying-in-the-interior-of-the-parallelogram-ABCD-two-parallels-to-AB-and-AD-are-drawn-intersecting-the-sides-of-ABCD-at-the-points-P-Q-R-S-See-Figure-Prove-that-M-lies-

Question Number 17645 by Tinkutara last updated on 09/Jul/17

Suppose that the point M lying in the

interior of the parallelogram A B C D,

two parallels to A B and A D are drawn,

intersecting the sides of A B C D at the

points P, Q, R, S (See Figure) . Prove

that M lies on the diagonal A C if and

only if [M R D S] = [M P B Q] .

Commented by Tinkutara last updated on 09/Jul/17

Answered by alex041103 last updated on 09/Jul/17

F G ∥ B C ∥ A D a n d E H ∥ A B ∥ C D

\Rightarrow A G M E a n d M H C F a r e p a r a l l e l o g r a m s

\Rightarrow S_{A M G} = S_{A M E} a n d S_{M C H} = S_{M C F}

A l s o A B C D i s p a r a l l e l o g r a m

\Rightarrow S_{A C D} = S_{A C B}

S_{A C D} = S_{A M C} + S_{D E M F} + S_{A M E} + S_{C F M}

a n d

S_{A C B} = S_{A M G} + S_{B H M G} + S_{C H M} - S_{A M C}

B u t S_{A M G} = S_{A M E} a n d S_{M C H} = S_{M C F}

a n d S_{A C D} = S_{A C B}

\Rightarrow S_{B H M G} = S_{D E M F} + 2 S_{A M C}

I f S_{B H M G} = S_{D E M F} t h e n S_{A M C} = 0

\Rightarrow S_{A M C} = \frac{d i s t a n c e (M t o A C) \times A C}{2} = 0

A n d A C \neq 0 t h e n d i s t a n c e (M t o A C) = 0

\Rightarrow M \in A C

Commented by alex041103 last updated on 09/Jul/17

Commented by Tinkutara last updated on 09/Jul/17

Thanks Sir!

Answered by ajfour last updated on 09/Jul/17

Commented by ajfour last updated on 09/Jul/17

Area MRDS = k \bar{a} \times (\bar{b} - m \bar{b})

= k (1 - m) \bar{a} \times \bar{b}

Area MPBQ = (\bar{a} - k \bar{a}) \times m \bar{b}

= m (1 - k) \bar{a} \times \bar{b}

position vector of M :

{\bar{r}}_{M} = k \bar{a} + m \bar{b}

when the two areas are equal,

k (1 - m) = m (1 - k)

\Rightarrow k = m

then {\bar{r}}_{M} = k (\bar{a} + \bar{b})

\Rightarrow M is on AC then .

Commented by Tinkutara last updated on 09/Jul/17

Thanks Sir!