Neka je γ = ∠ B C A \gamma=\angle BCA γ = ∠ B C A δ = ∠ B D C \delta=\angle BDC δ = ∠ B D C x = ∣ B D ∣ x=|BD| x = ∣ B D ∣
Prema poučku o kosinusu vrijedi
cos γ = 8 2 + 9 2 − 7 2 2 ⋅ 8 ⋅ 9 = 2 3 , sin γ = 1 − 4 9 = 5 3 . \cos\gamma=\frac{8^2+9^2-7^2}{2\cdot8\cdot9}=\frac23, \qquad \sin\gamma=\sqrt{1-\frac49}=\frac{\sqrt5}{3}. cos γ = 2 ⋅ 8 ⋅ 9 8 2 + 9 2 − 7 2 = 3 2 , sin γ = 1 − 9 4 = 3 5 . Prema poučku o sinusu vrijedi
x sin γ = 8 sin δ , \frac{x}{\sin\gamma}=\frac8{\sin\delta}, sin γ x = sin δ 8 , pa je
x = 8 sin γ sin δ = 8 sin γ sin ( 180 ∘ − ( 45 ∘ + γ ) ) = 8 sin γ sin ( 45 ∘ + γ ) x=\frac{8\sin\gamma}{\sin\delta} =\frac{8\sin\gamma}{\sin(180^\circ-(45^\circ+\gamma))} =\frac{8\sin\gamma}{\sin(45^\circ+\gamma)} x = sin δ 8 sin γ = sin ( 18 0 ∘ − ( 4 5 ∘ + γ )) 8 sin γ = sin ( 4 5 ∘ + γ ) 8 sin γ = 8 sin γ sin 45 ∘ cos γ + cos 45 ∘ sin γ = 8 ⋅ 5 / 3 ( 2 / 2 ) ⋅ ( 2 / 3 ) + ( 2 / 2 ) ⋅ ( 5 / 3 ) = 8 10 2 + 5 = 8 10 ( 5 − 2 ) . =\frac{8\sin\gamma}{\sin45^\circ\cos\gamma+\cos45^\circ\sin\gamma} =\frac{8\cdot\sqrt5/3}{(\sqrt2/2)\cdot(2/3)+(\sqrt2/2)\cdot(\sqrt5/3)} =\frac{8\sqrt{10}}{2+\sqrt5} =8\sqrt{10}(\sqrt5-2). = sin 4 5 ∘ cos γ + cos 4 5 ∘ sin γ 8 sin γ = ( 2 /2 ) ⋅ ( 2/3 ) + ( 2 /2 ) ⋅ ( 5 /3 ) 8 ⋅ 5 /3 = 2 + 5 8 10 = 8 10 ( 5 − 2 ) . Omjer površina je
P ( A B C ) P ( D B C ) = 8 ⋅ 9 sin γ x ⋅ 8 sin 45 ∘ = 9 ⋅ 5 / 3 8 10 ( 5 − 2 ) ⋅ 2 / 2 \frac{P(ABC)}{P(DBC)} =\frac{8\cdot9\sin\gamma}{x\cdot8\sin45^\circ} =\frac{9\cdot\sqrt5/3}{8\sqrt{10}(\sqrt5-2)\cdot\sqrt2/2} P ( D B C ) P ( A B C ) = x ⋅ 8 sin 4 5 ∘ 8 ⋅ 9 sin γ = 8 10 ( 5 − 2 ) ⋅ 2 /2 9 ⋅ 5 /3 odnosno
P ( A B C ) P ( D B C ) = 3 8 ( 5 − 2 ) = 3 ( 5 + 2 ) 8 . \frac{P(ABC)}{P(DBC)}=\frac3{8(\sqrt5-2)}=\frac{3(\sqrt5+2)}8. P ( D B C ) P ( A B C ) = 8 ( 5 − 2 ) 3 = 8 3 ( 5 + 2 ) . 
