In Fig. 15.86, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that
(i) ar(Δ BDE) = ar(Δ ABC)
(ii) ar(Δ BDE) = ar(Δ BAE)
(iii) ar(Δ BFE) = ar(Δ AFD)
(iv) ar(Δ ABC) = ar(Δ BEC)
(v) ar(Δ FED) = ar(Δ AFC)
(vi) ar(Δ BFE) = 2ar(Δ EFD)
Given that,
ABC and BDF are two equilateral triangles
Let,
AB = BC = CA = x
Then,
BD = = DE = BF
(i) We have,
Area ( = x2
Area ( = ()2
= * x2
Area ( = Area (
(ii) It is given that triangles ABC and BED are equilateral triangles
∠ACB = ∠DBE = 60o
BE ‖ AC (Since, alternate angles are equal)
Triangles BAF and BEC re on the same base BE and between the same parallels BE and AC
Therefore,
Area ( = Area (
Area ( = 2 Area ( (Therefore, ED is the median)
Area ( = Area (BAE)
(iii) Since,
are equilateral triangles
Therefore,
∠ABC = 60o and,
∠BDE = 60o
∠ABC = ∠BDE
AB ‖ DE
Triangles BED and AED are on the same base ED and between the same parallels AB and DE
Therefore,
Area ( = Area (
Area ( - Area ( = Area ( - Area (
Area ( = Area (
(iv) Since,
ED is the median of
Therefore,
Area ( = 2 Area (
Area ( = 2 * Area (
Area ( Area (
Area ( = 2 Area (
(v) Let h be the height of vertex E, corresponding to the side BD on
Let H be the vertex A, corresponding to the side BC in
From part (i), we have
Area ( = Area (
* BD * h = ( * BC * H)
= (2 BD * H)
h = H (1)
From part (iii), we have
Area ( = Area (
= * PD * H
= * FD * 2h
= 2 ( * FD * h)
= 2 Area (
(vi) Area ( = Area ( + Area (
= Area ( + Area (
[Using part (iii) and AD is the median of Area
= Area ( + * 4 Area ( [Using part (i)]
Area ( = 2 Area ( (2)
Area ( = Area ( + Area (
2 Area ( + Area (
3 Area ( (3)
From above equations,
Area (= 2 Area ( + 2 * 3 Area ()
= Area ()
Hence,
Area ( = Area (