Em um triângulo ABC de perimetro 30 cm, o lado BC mede 10 cm e a distância entre os pés das bissetrizes que partem de A é igual a 24cm. Calcule os lados AB e AC do triângulo.
Solução:
Sejam AC = y, CD = x,
![[;\angle CAD=\alpha;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vuxW3D9Sv-cCygCwORMFw8klnnZ31QhbRlKw4cKOrRPb1LY0XE3PJT13Y8GftRH-GSVLTS_sweFcWO98qgfvrcB-m7KPeQp2Ha8g_EEj4=s0-d "\angle CAD=\alpha")
e ![[;\angle BEA=\beta;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uoPeLD09Hl_IvzziNsWQ_aZAUZHzCRmxAq8NYivxcA5Nh-NBGcTkgoAl4CBzd_pUgnni5-WKDpwmYtwxYWLkZe_iaGaAV9kAN3GsS1Ixo=s0-d "\angle BEA=\beta")
Como AD é bissetriz interna e AE é bissetriz externa:
![[;\angle CAD=\angle BAD;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sg1wi-9Y-9MprQx62AW0PApwRWh8hgoyVvX28yzhkLN-Bf5Im_2AI75FCtbpDhnFY_jwQ2c_t-xL_wPBfPIld-BXU6Itp4urYzSIHnQzDu-SRHlB-f=s0-d "\angle CAD=\angle BAD")
![[;\angle FAE=\angle EAB;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uxKomYAKk0uZZvcXr92oB8LvuG7NNbyuuJdED1eDMHSFjlZU9UbcxU0PXefB_G19COZOT8pFPkduFo7aHiK_HAfSce8-APtHXY2ovvGUbQgdNe6z4=s0-d "\angle FAE=\angle EAB")
Logo:
![[;\angle EAD=90^{o};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vJI8K33eZ26hgrkUZxEOWe8ooih4k3rDuMoKZyJU2e8B75cIrGeHVXVKDofdTPDXCZWPDvxNLybd55p-o7PHakN5tefeDp0YnhC8yz9AiYHgrd=s0-d "\angle EAD=90^{o}")
Do teorema da bissetriz interna:
![[;\frac{x}{y}=\frac{20-y}{10-x};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vSJCbOB7nOkln0jyEtEKUqtaq7s7l4qzwegAlMfJByJP3qZcH4TcVtdI34fjNqNSz9OGxR3mbDRvpzEIvef2CxAe3NMJ7eQIO2ZZc3n4_AyJxnlwhV9a19XSNgp_eQ3J4IHUSOxVU7qO8=s0-d "\frac{x}{y}=\frac{20-y}{10-x}")
Aplicando o teorema dos senos:
![[;\Delta ACE:\ \frac{sin(90^{o}+\alpha)}{24+x}=\frac{sin\beta}{y}\Rightarrow \frac{sin\beta}{cos\alpha}=\frac{2x}{24+x};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_s5YCAlt03iqh-wmQpgYHl9REprD6sGFksy01BoUbBUf8Ug3U2eNJK0ivETqz-K45r78uibZeHw2qE-IU9LIxyVk69dad1byWA8A8CVbEfS1435CVxvXPtP8MZHhLokaT35l-iv2qqZoIXCUUpdhH48VS6nKF2JpBefnX5EKAszhfWM5h4ERS8IWidm7nWVdeVjIvhANBhJLX_MoGUTNVKhM1QO4bjYMOLfiqDZWcYREaf1sTAZZP5iVV8VTnbHRZm7Vy7E737MSaYGeTwNAAQhmO9A5_vwstdFOEA_YMyRBm7LoXMFveKBfl0w6tOMAdJxGiqL=s0-d "\Delta ACE:\ \frac{sin(90^{o}+\alpha)}{24+x}=\frac{sin\beta}{y}\Rightarrow \frac{sin\beta}{cos\alpha}=\frac{2x}{24+x}")
(I)![[;\Delta ABE:\ \frac{sin(90^{o}-\alpha)}{24-(10-x)}=\frac{sin\beta}{20-y}\Rightarrow \frac{sin\beta}{cos\alpha}=\frac{2(10-x)}{14+x};]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_ulqr9zwW_g-hH-yAVf43qRC5M4mSKzV9IMLTdVa5CjOaPr_meV2OQq2j0YigPMIus5eO6CvIkyP0mYRq5OkKznjakBcPU-ptbiqd751TF4LQTsBlPWjVmVQPQJXBaEU1Ph-D0kDCQNgPx9qgK0N88TzVrgTXqrVKR4XLmBa1laX0OWzwdhwbCALqcdtRfTviKN6tX1MpA6CogZw-4JcGC08KE5C32QyJOmbsze0d8JSi_WmZB_4O0Irs3F8JTiuTERTKa2FF6pfpoAlPG7V4y5VtKUzemb-oqeRN1MfS_yS_yO7OuF6K3A4PfwjZZbBCdd32_JifhBL7w71JtGi7bZJmJ8fgfmEZk=s0-d "\Delta ABE:\ \frac{sin(90^{o}-\alpha)}{24-(10-x)}=\frac{sin\beta}{20-y}\Rightarrow \frac{sin\beta}{cos\alpha}=\frac{2(10-x)}{14+x}")
(II)Igualando (I) e (II):
![[;\frac{2x}{24+x}=\frac{2(10-x)}{14+x}\Rightarrow x^{2}+14x-120=0;]](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_v6_HeGJ-IcrqlsUdDVh82p-zOkacpiV4WgQo7WyvzgKjQoaEQluhNZvdM2rlrFyuciplH0dzy4hP1GkMsRrCMg2JCtz1pk265fhAh2cfKSIJyijlWqo7cZe90gCTuHhpFdaFmmKeQcxSRL-45RW1Ih6_3snLh2BnRfj-6pgkF1JkhtPQYchZuo1aGSvjSu2XdnxKWa7iz4rvo=s0-d "\frac{2x}{24+x}=\frac{2(10-x)}{14+x}\Rightarrow x^{2}+14x-120=0")
Assim, x=6cm, AC= 12cm e AB=8cm.
Referência bibliográfica
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