Neka je a a a
∣ A E ∣ = ∣ B E ∣ = a 2 + ( a 2 ) 2 = a 5 2 |AE|=|BE|=\sqrt{a^2+\left(\frac a2\right)^2}=\frac{a\sqrt5}{2} ∣ A E ∣ = ∣ B E ∣ = a 2 + ( 2 a ) 2 = 2 a 5 P A B C D = 2 P A E B P_{ABCD}=2P_{AEB} P A B C D = 2 P A E B a 2 = ∣ A E ∣ ⋅ ∣ B F ∣ = a 5 2 ⋅ ∣ B F ∣ a^2=|AE|\cdot|BF|=\frac{a\sqrt5}{2}\cdot|BF| a 2 = ∣ A E ∣ ⋅ ∣ B F ∣ = 2 a 5 ⋅ ∣ B F ∣ ∣ B F ∣ = 2 a 5 5 . |BF|=\frac{2a\sqrt5}{5}. ∣ B F ∣ = 5 2 a 5 . ∣ E F ∣ = ∣ B E ∣ 2 − ∣ B F ∣ 2 = 3 a 5 10 . |EF|=\sqrt{|BE|^2-|BF|^2}=\frac{3a\sqrt5}{10}. ∣ E F ∣ = ∣ B E ∣ 2 − ∣ B F ∣ 2 = 10 3 a 5 . ∣ E F ∣ : ∣ F B ∣ : ∣ E B ∣ = 3 a 5 10 : 2 a 5 5 : a 5 2 |EF|:|FB|:|EB|=\frac{3a\sqrt5}{10}:\frac{2a\sqrt5}{5}:\frac{a\sqrt5}{2} ∣ E F ∣ : ∣ F B ∣ : ∣ E B ∣ = 10 3 a 5 : 5 2 a 5 : 2 a 5 ∣ E F ∣ : ∣ F B ∣ : ∣ E B ∣ = 3 : 4 : 5. |EF|:|FB|:|EB|=3:4:5. ∣ E F ∣ : ∣ F B ∣ : ∣ E B ∣ = 3 : 4 : 5. Ako je ∣ A E ∣ = ∣ B E ∣ |AE|=|BE| ∣ A E ∣ = ∣ B E ∣ d = d ( C , B E ) = 4 d=d(C,BE)=4 d = d ( C , B E ) = 4 2 P B C E = P A B E 2P_{BCE}=P_{ABE} 2 P B C E = P A B E
d ⋅ ∣ B E ∣ = ∣ B F ∣ ⋅ ∣ A E ∣ 2 , d\cdot|BE|=\frac{|BF|\cdot|AE|}{2}, d ⋅ ∣ B E ∣ = 2 ∣ B F ∣ ⋅ ∣ A E ∣ , 4 ⋅ ∣ B E ∣ = ∣ B F ∣ ⋅ ∣ B E ∣ 2 , 4\cdot|BE|=\frac{|BF|\cdot|BE|}{2}, 4 ⋅ ∣ B E ∣ = 2 ∣ B F ∣ ⋅ ∣ B E ∣ , ∣ B F ∣ = 8 cm . |BF|=8\ \text{cm}. ∣ B F ∣ = 8 cm . Tada iz
∣ B F ∣ = 2 a 5 5 = 8 |BF|=\frac{2a\sqrt5}{5}=8 ∣ B F ∣ = 5 2 a 5 = 8 slijedi a = 20 5 cm a=\frac{20}{\sqrt5}\ \text{cm} a = 5 20 cm a = 4 5 cm a=4\sqrt5\ \text{cm} a = 4 5 cm
