<多選題>
在 \(\triangle ABC\) 中,\(\overline{AB} = 6\),\(\overline{AC} = 5\),\(\overline{BC} = 4\)。令 \(\overline{AB}\) 中點為 \(D\),\(P\) 為 \(\angle ABC\) 之角平分線與 \(\overline{CD}\) 之交點,如右圖所示。試選出正確的選項。(1) \(\overline{CP} = \frac{3}{7}\overline{CD}\)
(2) \(\overset{\rightharpoonup}{AP} = \frac{3}{7}\overset{\rightharpoonup}{AB} + \frac{2}{7}\overset{\rightharpoonup}{AC}\)
(3) \(\cos \angle BAC = \frac{3}{4}\)
(4) \(\triangle ACP\) 面積為 \(\frac{15}{14}\sqrt{7}\)
(5) (內積) \(\overset{\rightharpoonup}{AP} \cdot \overset{\rightharpoonup}{AC} = \frac{120}{7}\)
答案
(1) ✗:\(\overline{CP}=\frac{4}{7}\overline{CD}\);(2) ✗:\(\overset{\rightharpoonup}{AP}=\frac{2}{7}\overset{\rightharpoonup}{AB}+\frac{3}{7}\overset{\rightharpoonup}{AC}\);
(3) ✓:餘弦定理得 \(\cos \angle BAC=\frac{3}{4}\);(4) ✓:面積計算得 \(\frac{15\sqrt{7}}{14}\);
(5) ✓:內積計算得 \(\frac{120}{7}\)。
故選(3)(4)(5)。 報錯
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