Data : ∆ABC and ∆BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC at F,
To Prove:
(i) ar.(∆BDE) = \(\frac{1}{4}\)ar.(∆ABC)
(ii) ar.(∆BDE) = \(\frac{1}{2}\)ar.(∆BAE)
(iii) ar.(∆ABC) = 2ar.(∆BEC)
(iv) ar.(∆BFE) = ar.(∆AFD
(v) ar.(∆BFE) = 2ar.(∆FE:
(vi) ar.(∆FED)= \(\frac{1}{8}\)ar.(∆AFC)
Construction: BC and AD are joined. G and H are midpoints of AB and AC resply. GH, GD and HD are joined.
Proof: (i) D is the mid point of BC.
Now, GH || BC OR GH || BD (Data)
GH = \(\frac{1}{2}\)BC OR GH = BD
∴ BDHG is a parallelogram.
∴ ar.(∆GHD) = ar. (∆BDG)
Similarly, GDCH is a parallelogrm.
∴ ar.(∆GHD) = ar.(∆DCH)
AGDH is a parllelogrm.
∴ ar.(∆GHD) = ar.(∆AGH)
∆ABC = ∆BDG + ∆GHD + ∆AGH + ∆HDC
∴ ∆ABC = 4 × ∆BDE
∴ ar.(∆BDE) = \(\frac{1}{4}\) × ar.(∆ABC).
(ii) ∆BDE = ∆AED
(∆BDE) – (∆FED) = (∆AED) – (∆FED)
∴ (∆BEF) = (∆ADF) ………. (i)
(∆ABD) = (∆ABF) + (∆ADF)
(∆ABD) = (∆ABF) + (∆BEF) [from (i)]
∴ (∆ABD) = (∆ABE) …………. (ii)
But, (∆ABD) = \(\frac{1}{2}\) × (∆ABC)
(∆ABD) = \(\frac{1}{2}\) × 4W(∆BDE)
∴ ar.(∆ABD) = 2 × ar.(∆BDE) ………… (iii)
From (i) and (ii), ar.(∆ABE) = 2 × ar.(∆BDE)
∴ ar.(∆BED) = \(\frac{1}{2}\) × ar.(∆ABE)
(iii) ∆ABE and ∆BEC are on same base BE and in between AC || BE.
∴ (∆ABE) = (∆BEC)
But, ar.(∆ABE) = 2 × ar.(∆BDE) (proved)
∴ 2ar.(∆BDE) = ar.(∆BEC)
But, ar.(∆BDE) = \(\frac{1}{4}\) × ar.(∆ABC)
∴ 2 × \(\frac{1}{4}\)ar.(∆ABC) = ar.(∆BEC)
∴ r.(∆BEC) = \(\frac{1}{2}\) × ar.(∆ABC)
ar.(∆ABC) = 2 × ar.(∆BEC)
(iv) ∆BED and ∆AED are on same base DE and in between AB || DE.
∴ ar. (∆BED) = ar.(∆AED)
ar.(∆BED) – ar.(∆DEF)
= ar.(∆AED) – ar.(∆DEF)
ar.(∆BEF) = ar.(∆AFD)
(v) Let height of ∆ABC be ‘a’ and height of DBDE be V.
(vi) ar.(∆AFC)= ar.(∆AFD) + ar.(∆ADC)
= ar. (∆BEF) + \(\frac{1}{2}\)ar.(∆ABC)
= ar.(∆BEF) + \(\frac{1}{2}\)× 4ar.(∆BED)
= ar.(∆BFE) + 2ar.(∆BDE)
But, ar.(∆BED) = ar.(∆BFE) + ar.(∆FED)
= 2 × ar.(∆FED) + ar.(∆FED) = 3 × ar.(∆FED)
∴ ar.(∆AFC) = 2 × ar.(∆FED) + 2 × 3 ar.(∆FED)
= 2 × ar.(∆FED) + 6 ×ar.(∆FED)
ar.(∆AFC) = 8 × ar.(∆FED)
ar.(∆FED) = \(\frac{1}{8}\)× ar.(∆AFC)
