∣ C K ∣ = ∣ K L ∣ = ∣ L D ∣ = 1 3 ∣ C D ∣ |CK|=|KL|=|LD|=\frac13|CD| ∣ C K ∣ = ∣ K L ∣ = ∣ L D ∣ = 3 1 ∣ C D ∣ ∣ K F ∣ = 1 2 ∣ K L ∣ = 1 2 ⋅ 1 3 ∣ C D ∣ = 1 6 ∣ C D ∣ |KF|=\frac12|KL|=\frac12\cdot\frac13|CD|=\frac16|CD| ∣ K F ∣ = 2 1 ∣ K L ∣ = 2 1 ⋅ 3 1 ∣ C D ∣ = 6 1 ∣ C D ∣ Kut B A E BAE B A E
∣ A E ∣ = ∣ B E ∣ 2 − ∣ A B ∣ 2 = 12 cm . |AE|=\sqrt{|BE|^2-|AB|^2}=12\text{ cm}. ∣ A E ∣ = ∣ B E ∣ 2 − ∣ A B ∣ 2 = 12 cm . Iz sličnosti trokuta A B E ABE A B E F S K FSK F S K
∣ A B ∣ ∣ A E ∣ = ∣ S F ∣ ∣ K F ∣ , 16 12 = ∣ S F ∣ 1 6 ∣ C D ∣ , ∣ S F ∣ = 2 9 ∣ C D ∣ . \frac{|AB|}{|AE|}=\frac{|SF|}{|KF|},\qquad\frac{16}{12}=\frac{|SF|}{\frac16|CD|},\qquad |SF|=\frac29|CD|. ∣ A E ∣ ∣ A B ∣ = ∣ K F ∣ ∣ S F ∣ , 12 16 = 6 1 ∣ C D ∣ ∣ S F ∣ , ∣ S F ∣ = 9 2 ∣ C D ∣. Iz pravokutnog trokuta D F S DFS D F S
( 2 9 ∣ C D ∣ ) 2 + ( ∣ C D ∣ 2 ) 2 = 100 \left(\frac29|CD|\right)^2+\left(\frac{|CD|}{2}\right)^2=100 ( 9 2 ∣ C D ∣ ) 2 + ( 2 ∣ C D ∣ ) 2 = 100 97 324 ∣ C D ∣ 2 = 100 \frac{97}{324}|CD|^2=100 324 97 ∣ C D ∣ 2 = 100 ∣ C D ∣ = 180 97 = 180 97 97 cm . |CD|=\frac{180}{\sqrt{97}}=\frac{180\sqrt{97}}{97}\text{ cm}. ∣ C D ∣ = 97 180 = 97 180 97 cm . Obodni kut α \alpha α C D CD C D ∠ C S D \angle CSD ∠ C S D
α = 1 2 ∠ C S D = ∠ F S D . \alpha=\frac12\angle CSD=\angle FSD. α = 2 1 ∠ C S D = ∠ F S D . Tada je
sin α = 1 2 ∣ C D ∣ r = 9 97 97 . \sin\alpha=\frac{\frac12|CD|}{r}=\frac{9\sqrt{97}}{97}. sin α = r 2 1 ∣ C D ∣ = 97 9 97 . 
