(c) x = \(\frac{bh}{2h+b}\)
Let AD be the height of the given triangle ABC, where AD = h and BC = b Also let height HG of rectangle EFGH equal \(x\) so that HE = GF = 2x
Now ΔBGH ~ ΔBDA
⇒ \(\frac{BG}{BD} =\frac{HG}{AD}\)
⇒ \(\frac{BG}{BD} =\frac{x}{h}\) ⇒ BG = \(\frac{x}{h}\)BD ....(i)
Also, ΔCFE ~ ΔCDA
⇒ \(\frac{Cf}{CD} =\frac{EF}{AD}\) = \(\frac{x}{h}\)
⇒ CF = \(\frac{x}{h}\)CD .....(ii)
BG + CF = BC – GF = b – 2x …(iii)
∴ From (i), (ii) and (iii)
b - 2x = \(\frac{x}{h}\) (BD + CD) ⇒ b - 2x = \(\frac{x}{h}\)BC = \(\frac{x}{h}\)b
⇒ bh – 2xh = xb ⇒ bh = xb + 2xh = x(b + 2h)
⇒ x = \(\frac{bh}{2h+b}\)