El arco capaz es el lugar geométrico de los puntos que unidos con los extremos de un segmento forman siempre un mismo ángulo.
Trazado del arco capaz
Vamos a realizar dos trazados del arco capaz, del ángulo
![\alpha \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vknv7CVwlgC2TdIpsaofg9ZecYL46MY1kRrM_bDKnQkSFTcS4c4h05brvFxHLpV1LfLBlBULhrg2HaB7ZzlLq1QfbzRDkFWc468ejqmugmygh9TxzyGUueuPKgPktXF8TEhnu-30INQ68s3wfTp5tHYlOj=s0-d)
para un segmento
![AB \ AB \](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGQEmLAB5SMq_Bx075li58BwlKOcLNoLsCFuU9y39na4py46TqajoPznBppGvnQaT2iP664pAnkX7_t7GnLcnAukw0QMOGQP6-lrToBSWpdVzaIb_ijuz2S-RIqXhBpdGLBUlldvlG5kh5P8UMYm47wXM=s0-d)
.
Trazado I:
Dibujamos la mediatriz del segmento
![AB \ AB \](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGQEmLAB5SMq_Bx075li58BwlKOcLNoLsCFuU9y39na4py46TqajoPznBppGvnQaT2iP664pAnkX7_t7GnLcnAukw0QMOGQP6-lrToBSWpdVzaIb_ijuz2S-RIqXhBpdGLBUlldvlG5kh5P8UMYm47wXM=s0-d)
, pues el centro del arco estará sobre ella, al ser equidistante de
![A A](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBsoM8QjzOQuv5rMXTpBzuPkFfQSS8J8hFYY2eFOduViy4KD0jT74gDVr7v2X-UOrXaF7sfs9GCswKgtgeOC9vSCn8Ix0v6Wk6K5qVlEIvbZyZSWn1NY8Zeq2FgzAiqUDHd2AWtlns3oxPrbV7Vdb94jo=s0-d)
y de
![B B](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t_xksWvcDsX_wV_NLJYR3NhBMvw2RlRB5FUX5xkewOslVe8YWViu_SSsDEbZxW2lpmu9QkyBtkfSkFh0ZF2QeRWrbfJVcXOQrPBS_DWeDrOY5rt4qLnIqJzigD34FNZs_y3q-KzcffKnrZZgSURvDdrMGb=s0-d)
.
Dibujamos el ángulo
![90^\circ - \alpha 90^\circ - \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uc4aNHZUynGB6c44guTZF6nzY1ebDd1mlLVQIa3C_C1ldlpJDYeUtywOeut6BsFQosRG_GpI4B0tJ0TgveVL3VwfTM6N2PzGK38joSoFAY5eXwLxV9H8-oMzHsVe1BYOdmTbXGeBCrjcBOBV0JpPBCSWxs=s0-d)
,complementario del dado, con vértice en
![A A](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBsoM8QjzOQuv5rMXTpBzuPkFfQSS8J8hFYY2eFOduViy4KD0jT74gDVr7v2X-UOrXaF7sfs9GCswKgtgeOC9vSCn8Ix0v6Wk6K5qVlEIvbZyZSWn1NY8Zeq2FgzAiqUDHd2AWtlns3oxPrbV7Vdb94jo=s0-d)
. El lado de este ángulo cortará a la mediatriz en el punto
![C C](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sEhb-hKnh-nxCmSVi1n8WnTHk073nig42mysAZAwCJyZ_ffZadcZNslFDz5NJMzaOpMe8JBqG7nR02t7tJ0QJ_OVthw7XPxeoHUIitVbnlZv0PlSrmUNai2J_nsH9GwSk4p1YUhoyIemSvsIG4KW0j2n7p=s0-d)
, centro del arco buscado.
Comprobamos que el ángulo central que abarca el arco
![AB \ AB \](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGQEmLAB5SMq_Bx075li58BwlKOcLNoLsCFuU9y39na4py46TqajoPznBppGvnQaT2iP664pAnkX7_t7GnLcnAukw0QMOGQP6-lrToBSWpdVzaIb_ijuz2S-RIqXhBpdGLBUlldvlG5kh5P8UMYm47wXM=s0-d)
mide
![\ 2 \alpha \ 2 \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u-8HqV-V6MgH99kvEi3wmIOE1dVFTnKY4TzRBtah2Tss2BrE9eiLS5nM6QO9GzARi8XAyc3nk_HF9Kb21IKwx9t74kBZSlTHf9PeU5yuQ_MdJ57PPcb8jhaSRxKAyUcf-7rd4foi1Tk9w9KWD2CYQWgMr8=s0-d)
, lo que indica que todos los ángulos inscritos que abarquen el mismo arco medirán
![\alpha \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vknv7CVwlgC2TdIpsaofg9ZecYL46MY1kRrM_bDKnQkSFTcS4c4h05brvFxHLpV1LfLBlBULhrg2HaB7ZzlLq1QfbzRDkFWc468ejqmugmygh9TxzyGUueuPKgPktXF8TEhnu-30INQ68s3wfTp5tHYlOj=s0-d)
.
Trazado II:
Dibujamos la mediatriz del segmento
![AB \ AB \](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGQEmLAB5SMq_Bx075li58BwlKOcLNoLsCFuU9y39na4py46TqajoPznBppGvnQaT2iP664pAnkX7_t7GnLcnAukw0QMOGQP6-lrToBSWpdVzaIb_ijuz2S-RIqXhBpdGLBUlldvlG5kh5P8UMYm47wXM=s0-d)
, pues el centro del arco estará sobre ella, al ser equidistante de
![A A](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBsoM8QjzOQuv5rMXTpBzuPkFfQSS8J8hFYY2eFOduViy4KD0jT74gDVr7v2X-UOrXaF7sfs9GCswKgtgeOC9vSCn8Ix0v6Wk6K5qVlEIvbZyZSWn1NY8Zeq2FgzAiqUDHd2AWtlns3oxPrbV7Vdb94jo=s0-d)
y de
![B B](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_t_xksWvcDsX_wV_NLJYR3NhBMvw2RlRB5FUX5xkewOslVe8YWViu_SSsDEbZxW2lpmu9QkyBtkfSkFh0ZF2QeRWrbfJVcXOQrPBS_DWeDrOY5rt4qLnIqJzigD34FNZs_y3q-KzcffKnrZZgSURvDdrMGb=s0-d)
.
Dibujamos el ángulo
![\alpha \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vknv7CVwlgC2TdIpsaofg9ZecYL46MY1kRrM_bDKnQkSFTcS4c4h05brvFxHLpV1LfLBlBULhrg2HaB7ZzlLq1QfbzRDkFWc468ejqmugmygh9TxzyGUueuPKgPktXF8TEhnu-30INQ68s3wfTp5tHYlOj=s0-d)
con vértice en
![A A](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBsoM8QjzOQuv5rMXTpBzuPkFfQSS8J8hFYY2eFOduViy4KD0jT74gDVr7v2X-UOrXaF7sfs9GCswKgtgeOC9vSCn8Ix0v6Wk6K5qVlEIvbZyZSWn1NY8Zeq2FgzAiqUDHd2AWtlns3oxPrbV7Vdb94jo=s0-d)
, como se ve en la figura. Trazamos por
![A A](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_uBsoM8QjzOQuv5rMXTpBzuPkFfQSS8J8hFYY2eFOduViy4KD0jT74gDVr7v2X-UOrXaF7sfs9GCswKgtgeOC9vSCn8Ix0v6Wk6K5qVlEIvbZyZSWn1NY8Zeq2FgzAiqUDHd2AWtlns3oxPrbV7Vdb94jo=s0-d)
la perpendicular al lado de dicho ángulo, que cortará a la mediatriz en el punto
![C C](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_sEhb-hKnh-nxCmSVi1n8WnTHk073nig42mysAZAwCJyZ_ffZadcZNslFDz5NJMzaOpMe8JBqG7nR02t7tJ0QJ_OVthw7XPxeoHUIitVbnlZv0PlSrmUNai2J_nsH9GwSk4p1YUhoyIemSvsIG4KW0j2n7p=s0-d)
, centro del arco buscado.
Comprobamos que el ángulo central que abarca el arco
![AB \ AB \](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vGQEmLAB5SMq_Bx075li58BwlKOcLNoLsCFuU9y39na4py46TqajoPznBppGvnQaT2iP664pAnkX7_t7GnLcnAukw0QMOGQP6-lrToBSWpdVzaIb_ijuz2S-RIqXhBpdGLBUlldvlG5kh5P8UMYm47wXM=s0-d)
mide
![\ 2 \alpha \ 2 \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_u-8HqV-V6MgH99kvEi3wmIOE1dVFTnKY4TzRBtah2Tss2BrE9eiLS5nM6QO9GzARi8XAyc3nk_HF9Kb21IKwx9t74kBZSlTHf9PeU5yuQ_MdJ57PPcb8jhaSRxKAyUcf-7rd4foi1Tk9w9KWD2CYQWgMr8=s0-d)
, lo que indica que todos los ángulos inscritos que abarquen el mismo arco medirán
![\alpha \alpha](https://lh3.googleusercontent.com/blogger_img_proxy/AEn0k_vknv7CVwlgC2TdIpsaofg9ZecYL46MY1kRrM_bDKnQkSFTcS4c4h05brvFxHLpV1LfLBlBULhrg2HaB7ZzlLq1QfbzRDkFWc468ejqmugmygh9TxzyGUueuPKgPktXF8TEhnu-30INQ68s3wfTp5tHYlOj=s0-d)
.