Notăm tg A 1 ^ = t \tg{\widehat{A_1}} = t tg A 1 = t tg B 1 ^ = u . \tg{\widehat{B_1}} = u. tg B 1 = u .
tg ( A ^ ) = 12 5 = 2 t 1 − t 2 ⇔ 6 t 2 + 5 t − 6 = 0 ⇒ t = I E E A = 2 3 ( 1 ) \tg(\widehat{A}) = \dfrac{12}{5} = \dfrac{2t}{1-t^2} \Leftrightarrow 6t^2+5t-6=0 \Rightarrow \boxed{t= \dfrac{IE}{EA} = \dfrac{2}{3}} \quad (1) tg ( A ) = 5 12 = 1 − t 2 2 t ⇔ 6 t 2 + 5 t − 6 = 0 ⇒ t = E A I E = 3 2 ( 1 )
tg ( B ^ ) = 12 9 = 2 u 1 − u 2 ⇔ 2 u 2 + 3 u − 2 = 0 ⇒ u = I E E B = 1 2 ( 2 ) \tg(\widehat{B}) = \dfrac{12}{9} = \dfrac{2u}{1-u^2} \Leftrightarrow 2u^2+3u-2=0 \Rightarrow \boxed{u = \dfrac{IE}{EB} = \dfrac{1}{2}} \quad (2) tg ( B ) = 9 12 = 1 − u 2 2 u ⇔ 2 u 2 + 3 u − 2 = 0 ⇒ u = EB I E = 2 1 ( 2 )
14 = E A + E B = ( 1 ) , ( 2 ) 3 ⋅ I E 2 + 2 ⋅ I E ⇒ I E = 4 ⇒ ( 1 ) A E = 6 ⇒ I ( 6 , 4 ) . 14=EA+EB \overset{(1),(2)}{=} \dfrac{3\cdot IE}{2} + 2 \cdot IE \Rightarrow \boxed{IE=4} \overset{(1)}{\Rightarrow} \boxed{AE=6} \Rightarrow \boxed{I(6,4)}. 14 = E A + EB = ( 1 ) , ( 2 ) 2 3 ⋅ I E + 2 ⋅ I E ⇒ I E = 4 ⇒ ( 1 ) A E = 6 ⇒ I ( 6 , 4 ) .
