Question

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

Answer 1

Given:   In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC.

To prove:  9AD2  = 7AB2 

Construction:  Draw AE ⊥ BC.

Proof :

In a ΔABC and ΔACE

AB = AC ( Given)

AE = AE ( common)

∠AEB = ∠AEC = 90°

∴ ΔABC ≅ ΔACE ( For RHS criterion)

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

BE = EC = BC / 2

In a right angled triangle ADE

AD2 = AE2 + DE2 ---------(1)

In a right angled triangle ABE

AB2 = AE2 + BE2 ---------(2)

From equ (1) and (2) we obtain

⇒ 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 ΔABC, AB = BC = CA)

⇒ AD2 = AB2 + AB2 / 36 – AB2 / 4

⇒ AD2 = (36AB2 + AB2– 9AB2) / 36

⇒ AD2 = (28AB2) / 36

⇒ AD2 = (7AB2) / 9

9AD2 = 7AB2 .

Answer 2

Given: ΔABC is an equilateral triangle. D is point on BC such that 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

Considering On Question;-

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 ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎

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‎Its proved!!!