在Youtube看到一題有趣的題目，於是就拿過來做一下了
\(在△ABC中，D為\overline {AB}中點，且\angle ADC=15^{\circ}，\angle CDB=45^{\circ}，如下圖。\)
\(請問\angle BCD=？\)
三角函數的單元，會這樣解給你看：
\(1^{\circ}\)
\(首先，既然知道D為\overline {AB}中點，那基本上可以確定△ADC=△BDC\)
\(此外，\angle BDC=\angle ACD+\angle CAD，故\angle CAD=30^{\circ}\)
\(根據正弦定理，\frac{\overline {AC}}{\sin135^{\circ}}=\frac{\overline {CD}}{\sin30^{\circ}}=\frac{\overline {AD}}{\sin15^{\circ}}\) → \(\frac{\overline {AC}}{\frac{\sqrt{2}}{2}}=\frac{\overline {CD}}{\frac{1}{2}}=\frac{\overline {AD}}{\frac{\sqrt{6}-\sqrt{2}}{4}}\)
\(可先得到\overline {AC}=(\sqrt{3}+1)\overline {AD} \overline {CD}=\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}\)
\(2^{\circ}\)
\(在△BDC中根據餘弦定理，\overline {BC}^{2}=\overline {DB}^{2}+\overline {DC}^{2}-2\cdot\overline {DB}\cdot\overline {DC}\cdot\cos\angle BDC\)
\(在這之中，\)
\(\overline {DC}=\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}，因此\overline {DC}^{2}=(2+\sqrt{3})\overline {AD}^{2}\)
\(2\cdot\overline {DB}\cdot\overline {DC}\cdot\cos\angle BDC=2\cdot\overline {AD}\cdot\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}\cdot\frac{\sqrt{2}}{2}=(\sqrt{3}+1)\overline {AD}^{2}\)
\(\therefore\overline {BC}^{2}=\overline {AD}^{2}+(2+\sqrt{3})\overline {AD}^{2}-(\sqrt{3}+1)\overline {AD}^{2}=2\overline {AD}^{2}\)
\(\overline {BC}=\overline {AD}\sqrt{2}\)
\(3^{\circ}\)
\(在△BDC中，再次透過餘弦定理，\cos\angle BCD=\frac{\overline {BC}^{2}+\overline {DC}^{2}-\overline {BD}^{2}}{2\cdot\overline {BC}\cdot\overline {DC}}\)
\(在這之中，\)
\(\overline {BD}=\overline {AD}\)
\(\overline {BC}=\overline {AD}\sqrt{2}\)
\(\overline {DC}=\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}\)
\(將全部的線段以\overline {AD}代換後變成：\)
\(\cos\angle BCD=\frac{(\overline {AD}\sqrt{2})^{2}+(\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD})^{2}-\overline {AD}^{2}}{2\cdot(\overline {AD}\sqrt{2})\cdot(\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD})}=\frac{3+\sqrt{3}}{2\sqrt{3}+2}=\frac{3+\sqrt{3}}{2\sqrt{3}+2}=\frac{(\sqrt{3}+1)\cdot\sqrt{3}}{(\sqrt{3}+1)\cdot2}=\frac{\sqrt{3}}{2}\)
\(\therefore\angle BCD=30^{\circ}\)#
但是正常來說，這種做法很複雜，且容易計算錯誤。 
\(尺規作圖，找一點P使得\overline {DA}=\overline {DP}\)
\(由於\angle BDC=\angle ACD+\angle CAD，得\angle CAD=\angle PAD=30^{\circ}\)
\(在△DPA中，因\overline {DA}=\overline {DP}，故\angle DPA=\angle PAD=30^{\circ}\)
\(因此推得\angle PDC=135^{\circ}-\angle PDA=135^{\circ}-\frac{180^{\circ}-30^{\circ}\times2}{2}=15^{\circ}\)
\(在△DPC中，\angle PCD=\angle PDC=15^{\circ}，因此推得\overline {PD}=\overline {PC}\)
\(在△PDB中，\overline {PD}=\overline {DB}(=\overline {DA})，\angle PDB=\angle BDC+\angle PDC=60^{\circ}，\)
\(因此推得△PDB為正△，→\overline {PB}=\overline {PD}(=\overline {PC})\)
\(在△BPC中，\overline {BP}=\overline {CP}，\angle BPC=180^{\circ}-\angle BPD+\angle DPA=90^{\circ}，\)
\(因此推得\angle BCP=\angle CBP=\frac{180^{\circ}-90^{\circ}}{2}=45^{\circ}\)
\(\therefore\angle BCD=\angle BCP-\angle ADC=45^{\circ}-15^{\circ}=30^{\circ}\)#
\(首先，既然知道D為\overline {AB}中點，那基本上可以確定△ADC=△BDC\)
\(此外，\angle BDC=\angle ACD+\angle CAD，故\angle CAD=30^{\circ}\)
\(根據正弦定理，\frac{\overline {AC}}{\sin135^{\circ}}=\frac{\overline {CD}}{\sin30^{\circ}}=\frac{\overline {AD}}{\sin15^{\circ}}\) → \(\frac{\overline {AC}}{\frac{\sqrt{2}}{2}}=\frac{\overline {CD}}{\frac{1}{2}}=\frac{\overline {AD}}{\frac{\sqrt{6}-\sqrt{2}}{4}}\)
\(可先得到\overline {AC}=(\sqrt{3}+1)\overline {AD} \overline {CD}=\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}\)
\(2^{\circ}\)
\(在△BDC中根據餘弦定理，\overline {BC}^{2}=\overline {DB}^{2}+\overline {DC}^{2}-2\cdot\overline {DB}\cdot\overline {DC}\cdot\cos\angle BDC\)
\(在這之中，\)
\(\overline {DC}=\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}，因此\overline {DC}^{2}=(2+\sqrt{3})\overline {AD}^{2}\)
\(2\cdot\overline {DB}\cdot\overline {DC}\cdot\cos\angle BDC=2\cdot\overline {AD}\cdot\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}\cdot\frac{\sqrt{2}}{2}=(\sqrt{3}+1)\overline {AD}^{2}\)
\(\therefore\overline {BC}^{2}=\overline {AD}^{2}+(2+\sqrt{3})\overline {AD}^{2}-(\sqrt{3}+1)\overline {AD}^{2}=2\overline {AD}^{2}\)
\(\overline {BC}=\overline {AD}\sqrt{2}\)
\(3^{\circ}\)
\(在△BDC中，再次透過餘弦定理，\cos\angle BCD=\frac{\overline {BC}^{2}+\overline {DC}^{2}-\overline {BD}^{2}}{2\cdot\overline {BC}\cdot\overline {DC}}\)
\(在這之中，\)
\(\overline {BD}=\overline {AD}\)
\(\overline {BC}=\overline {AD}\sqrt{2}\)
\(\overline {DC}=\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD}\)
\(將全部的線段以\overline {AD}代換後變成：\)
\(\cos\angle BCD=\frac{(\overline {AD}\sqrt{2})^{2}+(\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD})^{2}-\overline {AD}^{2}}{2\cdot(\overline {AD}\sqrt{2})\cdot(\frac{\sqrt{6}+\sqrt{2}}{2}\overline {AD})}=\frac{3+\sqrt{3}}{2\sqrt{3}+2}=\frac{3+\sqrt{3}}{2\sqrt{3}+2}=\frac{(\sqrt{3}+1)\cdot\sqrt{3}}{(\sqrt{3}+1)\cdot2}=\frac{\sqrt{3}}{2}\)
\(\therefore\angle BCD=30^{\circ}\)#
但是正常來說，這種做法很複雜，且容易計算錯誤。
補習班老師會這樣教你(?)：
\(尺規作圖，找一點P使得\overline {DA}=\overline {DP}\)
\(由於\angle BDC=\angle ACD+\angle CAD，得\angle CAD=\angle PAD=30^{\circ}\)
\(在△DPA中，因\overline {DA}=\overline {DP}，故\angle DPA=\angle PAD=30^{\circ}\)
\(因此推得\angle PDC=135^{\circ}-\angle PDA=135^{\circ}-\frac{180^{\circ}-30^{\circ}\times2}{2}=15^{\circ}\)
\(在△DPC中，\angle PCD=\angle PDC=15^{\circ}，因此推得\overline {PD}=\overline {PC}\)
\(在△PDB中，\overline {PD}=\overline {DB}(=\overline {DA})，\angle PDB=\angle BDC+\angle PDC=60^{\circ}，\)
\(因此推得△PDB為正△，→\overline {PB}=\overline {PD}(=\overline {PC})\)
\(在△BPC中，\overline {BP}=\overline {CP}，\angle BPC=180^{\circ}-\angle BPD+\angle DPA=90^{\circ}，\)
\(因此推得\angle BCP=\angle CBP=\frac{180^{\circ}-90^{\circ}}{2}=45^{\circ}\)
\(\therefore\angle BCD=\angle BCP-\angle ADC=45^{\circ}-15^{\circ}=30^{\circ}\)#
