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Theorem 1: Parallelograms

X and Y are p...

Question

X and Y are points on the side LN of the triangle LMN such that LX=XY=YN.Through X, a line is drawn parallel to LM to meet MN at Z. Prove that ar(LZY)=ar(MZYX).

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Solution

$\begin{matrix} L X = X Y = Y N (g i v e n) ∵ L M ∥ X Z \Rightarrow D r a w Z L ⊥_{r} L Y \Rightarrow I n Δ L Z Y X T ∥ Y Z \Rightarrow a r Δ L Z Y = \frac{1}{2} (L Y) X Z L = (2 X Y) \times \frac{1}{2} \times (Z L) = (Z L) \times (X Y) \Rightarrow I n q u a d r i l a t e r a l (M Z Y X) = (b a s e \times h e i g h t) \Rightarrow a r (M Z Y X) = (X Y) \times (Z L) \Rightarrow H e n c e, a r (Δ L Z Y) = a r M Z Y X p r o v e d . \end{matrix}$

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Q. Question 2
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (\(\Delta\) LZY) = ar (MZYX).

Q. Question 2
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (

Δ

LZY) = ar (MZYX).

Q. X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (See figure). Then:

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Theorem 1: Parallelograms

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