in an equilateral triangle ABC D is a point on side BC such that BD is equal to 1 by 3 BC prove that 9ADsquare is equal to 7ABsquare
Answers
Answer:
I get AD sq=7AB sq
Step-by-step explanation:
Mark a point E at midway on BC.
Say triangle ABC has sides 3 3 and 3
So BD=1
DC=2
and
BE=1.5
AE is perp' to BC and also
BE=EC (as tri ABC is equil')
So
AB sq=AE sq + BE sq
= AE sq + 1.5*1.5
=2.1*2.1+1.5*1.5
=4.41+2.25=6.66
so 7AB sq
=7*6.66
=4.66
and
AD sq = AE sq + DE sq
= AE sq + 0.5*0.5
=2.1*2.1+0.25
=4.41 + 0.25
=4.66
=7AB sq.... ANSWER
but 9AD sq
=9*4.66
=41.94 is not equal to 7AB sq
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