In an equilateral triangle ABC, the side BC is trisected at P. Prove that 9AP²= 7BC²
Answers
Figure is in the attachment...
∆ABC is an equilateral triangle. BC is trisected at P.
Construction
- Draw AE perpendicular to BC
- Join A, P.
We have to prove ; 9AP²= 7BC²
Proof
As BC is trisected at P, that means BC = BP + PE + EC
So, BP = BC/3 ------ (i)
Now in ∆ABE & ∆ACE :
AB = AC [Equilateral triangle ; 3 sides are equal]
AE = AE [Common side]
∠AEB = ∠AEC [AE ⊥ BC]
So, ∆ABE ≅ ∆ACE [Side - Side - Angle theorem]
So, BE = CE
→ BE = CE = BC/2......(1)
Now in ∆APE ;
→ AP² = AE² + PE² ------ (2)
In ∆ABE ;
→ AB² = AE² + BE² ------ (3)
Now subtracting (3) from (2) :-
→ AP² - AB² = AE² + PE² - AE² - BE²
→ AP² - AB² = PE² - BE²
→ AP² - BC² = (BE - BP)² - (BC/2)²
→ AP² = BC² + [(BC/2) - (BC/3)]² - (BC/2)²
→ AP² = BC² + [(3BC - 2BC)/6]² - (BC²/4)
→ AP² = BC² + (BC²/36 - BC²/4)
→ AP² = BC² + [(BC² - 9BC²)/36]
→ AP² = BC² + (-8BC²/36)
→ AP² = BC² + (-2BC²/9)
→ AP² = (9BC² - 2BC² )/9
→ AP² = 7BC²/9
→ 9AP² = 7BC² [Proved]
Step-by-step explanation:
Figure is in the attachment...
∆ABC is an equilateral triangle. BC is trisected at P.
Construction
Draw AE perpendicular to BC
Join A, P.
We have to prove ; 9AP²= 7BC²
Proof
As BC is trisected at P, that means BC = BP + PE + EC
So, BP = BC/3 ------ (i)
Now in ∆ABE & ∆ACE :
AB = AC [Equilateral triangle ; 3 sides are equal]
AE = AE [Common side]
∠AEB = ∠AEC [AE ⊥ BC]
So, ∆ABE ≅ ∆ACE [Side - Side - Angle theorem]
So, BE = CE
→ BE = CE = BC/2......(1)
Now in ∆APE ;
→ AP² = AE² + PE² ------ (2)
In ∆ABE ;
→ AB² = AE² + BE² ------ (3)
Now subtracting (3) from (2) :-
→ AP² - AB² = AE² + PE² - AE² - BE²
→ AP² - AB² = PE² - BE²
→ AP² - BC² = (BE - BP)² - (BC/2)²
→ AP² = BC² + [(BC/2) - (BC/3)]² - (BC/2)²
→ AP² = BC² + [(3BC - 2BC)/6]² - (BC²/4)
→ AP² = BC² + (BC²/36 - BC²/4)
→ AP² = BC² + [(BC² - 9BC²)/36]
→ AP² = BC² + (-8BC²/36)
→ AP² = BC² + (-2BC²/9)
→ AP² = (9BC² - 2BC² )/9
→ AP² = 7BC²/9
→ 9AP² = 7BC² [Proved]