Question

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

Answer 1

First draw a perpendicular on BC with point E

Then with Pythagoras theorem u get:

AE^2+BE^2=AB^2

which can also be written as

AE^2+(AB/2)^2=AB^2

From which u find value of AE^2,I.e.

AE^2=3/4AB^2

Now come to triangleADE and apply Pythagoras theorem here

AE^2+DE^2=AD^2

Now as can see DE=AB/6 so put this value here of DE and also put value of AE^2 here to get:

3/4AB^2+AB^2/6=AD^2

By solving u will get:

28/36AB^2=AD^2

Further u will get your result:

7AB^2=9AD^2

Answer 2

Answer:

Step-by-step explanation:

Here is your solution

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 ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎proved

Hope it helps you