Question

*Hi, Mate*


Question:



In an EQUILATERAL triangle ABC, D is a point on side BC such that BD =1/3 BC. PROVED that 9AD^2=7AB^2.


Give Correct Answer.

Please

Answer 1

HEY THERE!!

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◀ 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 Creition

∴ ΔABC ≅ ΔACE

Again,

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

BE = EC = BC

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

 ◀ 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎‎
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‎Hence, 9AD² = 7AB² ‎‎‎‎‎‎‎‎‎‎‎‎‎
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‎Its proved!!!

☔ ✍✍✍✍✍✍✍✍✍✍✍✍✍✍✍ ☔

Answer 2

hey mate
here's the proof