Explicație pas cu pas:
a) ducem CN ⊥ AB
[tex]AC \: bisectoare => ∠DAC = ∠CAB = 45°[/tex]
[tex]=> ∠ACN = 45°[/tex]
[tex]=> ΔACN \: dreptunghic \: isoscel[/tex]
[tex]=> AN = CN[/tex]
[tex]AC = 5 \sqrt{2} \: cm => CN = 5 \: cm[/tex]
[tex]ΔCNB: NB² = BC² - CN² = 13² - 25² = 169 - 25 = 144[/tex]
[tex]=> NB = 12 cm[/tex]
[tex]cos(∠NBC) = \frac{NB}{BC} = \frac{12}{13}[/tex]
[tex]=> cos(∠ABC) = \frac{12}{13} [/tex]
b) M simetric
[tex]=> AC = CM = 5 \sqrt{2} \: cm[/tex]
[tex]DC = AN = 5 \: cm[/tex]
[tex]∠DCM = 90° + 45° = 135°[/tex]
[tex]cos135° = cos(90° + 45°) = - cos45° = - \frac{ \sqrt{2} }{2} [/tex]
[tex]ΔDCM: \: DM² = DC² + CM² - 2×DC×CM× cos(∠DCM) \: = {5}^{2} + ({5 \sqrt{2} })^{2} - 2 \times 5 \times 5 \sqrt{2} \times \cos(135) = 25 + 50 - 50 \sqrt{2} \times ( - \frac{ \sqrt{2} }{2}) = 75 + 50 = 125[/tex]
[tex]DM = \sqrt{125} = 5 \sqrt{5} \: cm[/tex]
