Question

in triangle abc is an equilateral triangle whose each sides x units P and Q are two points on BC produced such that AB=BC=CD prove that PQ/PA =PA/PB and PA^2 =3x^2

Answer 1

your question is incorrect. A correct question is ------> In the figure ΔABC is an equilateral triangle, whose each side measures x units. P and Q are two points on BC produced such that PB=BC=CQ
Prove that :
(a) PQ/PA = PA/PB
(b)PA²= 3x²

solution :- see figure, Draw AL perpendicular to BC.

in ∆ALB and ∆ALC ,
AL is common side of both triangles.
AB = AC [ ∆ABC is an equilateral triangle]
angle ALB = angle ALC = 90°
from RHS congruence rule,
∆ALB$\cong$∆ALC
so, BL = LC = BC/2 = x/2 [ as side length of ∆ABC is x]

now, from Pythagoras theorem,
∆ALC , AC² = AL² + LC²
AL² = AC² - LC² = x² - (x/2)² = 3x²/4
AL = √3x/2 ..........(1)

in ∆ALP,
PA² = PL² + AL²
= (PB + BL)² + (√3x/2)² [ from equation (1) ]
= (x + x/2)² + 3x²/4
= 9x²/4 + 3x²/4 = 12x²/4
PA² = 3x² [ hence proved ] .......(2)

again, PQ × PB = (PB + BC + CQ) × PB
= (x + x + x) × x
= 3x²
= PA² [from equation (2), ]
PQ × PB = PA²
PQ/PA = PA/PB [hence proved]

Answer 2

In ΔALB = ΔALC

∠ALB = ∠ALC (90° Each)

AB = AC (each equal to x)

AL = AL (Common)

ΔALB ≅ ΔALC (RHS)

= BL = LC = x/2

Now AL² = AC² - LC²

= x² - (x/2)²

= x² - x²/4

= 3x²/4

AL = √3x / 2

in ΔALP

PA² = PL² + AL²

PA² = (3x/2)² + (√3x/2)²

= 9x²/4 + 3x²/4

= 12x²/4

= 3x²

Now PQ x PB

= (PB + BC + CQ) x PB = (3x) . x

= 3x²

So PQ x PB = PA²

= PQ/PA

= PA/PB