Question

ΔABC is an equilateral triangle.<br />Poiny D is on BC such that,BD =1/5 BC<br />then prove that 25AD^=21AB^​

Answer 1

Answer:

Step-by-step explanation:

Given: In an equilateral triangle

ΔABC. The side BC is trisected at D

such that BD = (1/3) BC.

To prove: 9AD^2 = 7AB^2

Construction: Draw AE ⊥ BC.

Proof :

As height of an equilateral triangle

divides BC in equal ratio. BE = 1/2 BC

In a right angled triangle ADE

AD^2 = AE^2 + DE^2

AE^2 = AD^2 - DE^2 ---------(1)

In a right angled triangle ABE

AB^2 = AE^2 + BE^2

AE^2 = AB^2 - BE^2 ---------(2)

From equation (1) and (2) we obtain

》 AD^2 - AB^2 = DE^2 - BE^2 .

》 AD^2 - AB^2 = (BE – BD)^2 - BE^2 .

》AD^2 - AB^2 = (BC / 2 – BC/3)^2 – (BC/2)^2

》 AD^2 - AB^2 = ((3BC – 2BC)/6)^2 – (BC/2)^2

》 AD^2 - AB^2 = BC^2 / 36 – BC^2 / 4

( In a equilateral triangle ΔABC, AB = BC = CA)

》 AD^2 = AB^2 + AB^2 / 36 – AB^2 / 4

》 AD^2 = (36AB^2 + AB^2– 9AB^2) / 36

》 AD^2 = (28AB^2) / 36

》 AD^2 = (7AB^2) / 9

9AD^2 = 7AB^2