Question

Can someone please please help me please

Answer 1

Let:

DF = a, GF = b

DH = HG = x, FH = y

∠DHF = θ, ∠GHF = 180° - θ

Use Law of Cosines:

$a^2=x^2+y^2-2xy\cos\theta\\\\b^2=x^2+y^2-2xy\cos(180^o-\theta)\\\\b^2=x^2+y^2-2xy(-\cos\theta)\\\\b^2=x^2+y^2+2xy\cos\theta$

θ is in Quadrant II, therefore cosθ < 0.

Conclusion:

$2xy\cos\theta < 0,\ then:$

$x^2+y^2-2xy\cos\theta > x^2+y^2+2xy\cos\theta\Rightarrow a^2 > b^2\to a > b\\\\\boxed{DF > GF}$