Cho tứ giác ABCD có là góc nhọn tạo bởi hai đường chéo chứng minh rằng

Ta có: sin α = sin (180° − α)

S_ABCD = S_IAB + S_IBC + S_ICD + S_IDA

\( = \frac{1}{2}IA\,.\,IB\,.\,\sin \widehat {AIB} + \frac{1}{2}IB\,.\,IC\,.\,\sin \widehat {BIC} + \frac{1}{2}IC\,.\,ID\,.\,\sin \widehat {CID} + \frac{1}{2}ID\,.\,IA\,.\,\sin \widehat {DIA}\)

\( = \frac{1}{2}IA\,.\,IB\,.\,\sin \alpha + \frac{1}{2}IB\,.\,IC\,.\,\sin \alpha + \frac{1}{2}IC\,.\,ID\,.\,\sin \alpha + \frac{1}{2}ID\,.\,IA\,.\,\sin \alpha \)

\( = \frac{1}{2}\sin \alpha \left( {IA\,.\,IB\, + IB\,.\,IC\, + IC\,.\,ID\, + ID\,.\,IA} \right)\)

\[ = \frac{1}{2}\sin \alpha \left[ {IB\,\left( {IA\, + IC} \right)\, + ID\,\left( {IA\, + IC} \right)\,} \right]\]

\[ = \frac{1}{2}\sin \alpha \left( {IB + ID} \right)\left( {IA\, + IC} \right)\]

\[ = \frac{1}{2}AC\,.\,BD\,.\,\sin \alpha \] (đpcm)