Question

In an equilateral triangle ABC, D is a point on the side BC such that 1 BD BC. 3  Prove that 9AD2 = 7AB2

Answer 1

In equilateral∆ ABC . Given D is a point on side BC. BD= ⅓(BC)(GIVEN)
Draw a line AE Perpendicular to BC.
In a ΔABE and ΔACE
AB = AC ( Given) (equilateral∆)
AE = AE ( common)
∠AEB = ∠AEC = 90°
ΔABC ≅ ΔACE ( by RHS criterion)
BE = EC (By C.P.C.T)
BE = EC = BC / 2
BD= ⅓(BC)(GIVEN)
DE=BE-BD
In a right angled ∆ ADE
AD² = AE² + DE² ---------(1)
In a right angled ∆ ABE
AB²= AE² + BE² ---------(2)
From eq (1) and (2)
AD² - AB² = DE² - BE².
AD² - AB² = (BE – BD)² - BE² .
AD² - AB² = (BC/2 – BC/3)² – (BC/2)²
AD² - AB² = ((3BC – 2BC)/6)² – (BC/2)2
AD² - AB² = BC² / 36 – BC² / 4
AD²= AB² + AB² / 36 - AB² / 4
[In a equilateral triangle ΔABC, AB = BC = CA]
AD² = (36AB² + AB²– 9AB²) / 36
AD²= (28AB²)/36
AD² = (7AB²) / 9
9AD² = 7AB²

HOPE THIS WILL HELP YOU.....

Answer 2

Here is your solution

Given:-

ABC is an equilateral triangle.

D is point on BC .

so BD =BC.

To prove:-

 9 AD² = 7 AB²

Construction: Draw AE ⊥ BC.

Proof ;-

Considering on Triangles which are given below;-

In a ΔABC and ΔACE

AB = AC ( given)

AE = AE (common)

∠AEB = ∠AEC = (Right angle)

∴ ΔABC ≅ ΔACE 

By RHS Creation

∴ ΔABC ≅ ΔACE 

Again,

BE = EC (By C.P.C.T)

BE = EC = BC²

In a right angled ΔADE

AD²= AE2 + DE² ---(1)

In a right angled ΔABE

AB² = AE² + BE² ---(2)

From equation (1) and (2) ;

 =) AD²  - AB² =  DE² - BE².

 =) AD²  - AB² = (BE – BD)² - BE².

 = ) AD²  - AB² = (BC / 2 – BC/3)² – (BC/2)²

 = AD2  - AB2 = ((3BC – 2BC/6)² – (BC/2)² 

 = AD²  - AB² = (BC² / 36 – BC2 / 4 )

( In a equilateral triangle, All sides are equal to each other)

AB = BC = AC

 = ) AD²= AB² + AB²/ 36 – AB² / 4

 = )AD² = (36AB² + AB²– 9AB²) / 36

 = ) AD² = (28AB²) / 36

=) AD² = (7AB²) / 9

Cross Multiplication here,

= ) 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎

‎

‎

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‎Hence,

9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎proved

Hope it helps you