Question

in an equilateral triangle ABC D is a point on side BC such that BD is equal to one third of BC prove that 9ADsquare is equal to 7 AB square

Answer 1

Sol:
Given: In an equilateral triangle ΔABC. The side BC is trisected at D such that BD = (1/3) BC.
To prove: 9AD = 7AB
Construction: Draw AE ⊥ BC.
Proof :
In a ΔABC and ΔACE
AB = AC ( Given)
AE = AE ( common)
∠AEB = ∠AEC = 90°
∴ ΔABC ≅ ΔACE ( For RHS criterion)
BE = EC (By C.P.C.T)
BE = EC = BC / 2
In a right angled triangle ADE
AD = AE + DE ---------(1)
In a right angled triangle ABE
AB = AE + BE ---------(2)
From equ (1) and (2) we obtain
⇒ AD - AB = DE - BE .
⇒ AD - AB = (BE – BD) - BE .
⇒ AD - AB = (BC / 2 – BC/3) – (BC/2)
⇒ AD - AB = ((3BC – 2BC)/6) – (BC/2)
⇒ AD - AB = BC / 36 – BC / 4 ( In a equilateral triangle ΔABC, AB = BC = CA)
⇒ AD = AB + AB / 36 – AB / 4
⇒ AD = (36AB + AB – 9AB ) / 36
⇒ AD = (28AB ) / 36
⇒ AD = (7AB ) / 9
9AD = 7AB .

Answer 2

Answer is attached.

Here's is Answer key for whole maths Leaked board paper 2018 =>
brainly.in/question/3121329