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The Mid-Point Theorem

Question 5 iI...

Question

Question 5 (i)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that $a r (B D E) = \frac{1}{4} a r (A B C)$

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Solution

(i) Let G and H be the mid-points of side AB and AC respectively.
Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC. (Mid-point theorem)

$∴ G H = \frac{1}{2} B C a n d G H | | B D$
$\Rightarrow G H = B D = D C a n d G H | | B D$ (D is the mid-point of BC)
Consider quadrilateral GHDB.
GH || BD and GH = BD

Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.
Therefore, BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram.
We know that, in a parallelogram, the diagonal bisects it into two triangles of equal area.
Hence, $A r e a (Δ B D G) = A r e a (Δ H G D)$

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.
$a r (Δ G D H) = a r (Δ C H D)$ (For parallelogram DCHG)
$a r (Δ G D H) = a r (Δ H A G)$ (For parallelogram GDHA)
$a r (Δ B D E) = a r (Δ D B G)$ (For parallelogram BEDG)
$a r (Δ A B C) = a r (Δ B D G) + a r (Δ G D H) + a r (Δ D C H) + a r (Δ A G H)$
$a r (Δ A B C) = 4 \times a r (Δ B D E)$
Hence, $a r (B D E) = \frac{1}{4} a r (A B C)$

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

a r (B D E) = \frac{1}{2} a r (B A E)

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