In an equilateral triangle ABC,D is the point on the side BC such that BD =1/3BC prove that 9ADsquare =7ABsquare
Answers
Step-by-step explanation:
ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .
Now, ∆ABE and ∆AEC
∠AEB = ∠ACE = 90°
AE is common side of both triangles ,
AB = AC [ all sides of equilateral triangle are equal ]
From R - H - S congruence rule ,
∆ABE ≡ ∆ACE
∴ BE = EC = BC/2
Now, from Pythagoras theorem ,
∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)
∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)
From equation (1) and (2)
AB² - AD² = BE² - DE²
= (BC/2)² - (BE - BD)²
= BC²/4 - {(BC/2) - (BC/3)}²
= BC²/4 - (BC/6)²
= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9
∵AB = BC = CA
So, AB² = AD² + 2AB²/9
9AB² - 2AB² = 9AD²
Hence, 9AD² = 7AB²
hope it helps you.....
mark it as brainliest
Step-by-step explanation:
Given :-
A ∆ABC in which AB = BC = CA and D is a point on BC such that BD = ⅓BC.
To prove :-
9AD² = 7AB² .
Construction :-
Draw AL ⊥ BC .
Proof :-
In right triangles ALB and ALC, we have
AB = AC ( given ) and AL = AL ( common )
∴ ∆ALB ≅ ∆ ALC [ By RHS axiom ] .
So, BL = CL .
Thus, BD = ⅓BC and BL = ½BC .
In ∆ALB, ∠ALB = 90° .
∴ AB² = AL² + BL² .......(1) [ by Pythagoras' theorem ] .
In ∆ALD , ∠ALD = 90° .
∴ AD² = AL² + DL² . [ by Pythagoras' theorem ] .
⇒ AD² = AL² + ( BL - BD )² .
⇒ AD² = AL² + BL² + BD² - 2BL.BD .
⇒ AD² = ( AL² + BL² ) + BD² - 2BL.BD .
⇒ AD² = AB² + BD² - 2BL.BD. [ using (1) ]
⇒ AD² = BC² + ( ⅓BC )² - 2( ½BC ). ⅓BC .
[ ∵ AB = BC, BD = ⅓BC and BL = ½BC ] .
⇒ AD² = BC² + 1/9BC² - ⅓BC² .
⇒ AD² = 7/9BC² .
⇒ AD² = 7/9AB² [ ∵ BC = AB ] .
⇒ 9 AD² = 7 AB²
Hence proved