Question

[Toại]

Tứ giác ABCD có 2 đường chéo cắt nhau tại O .Cứng minh rằng diện tích ABCD = (1/2).AC.BD.sinAOB

Answer 1

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_Hình hơi xấu , thông cảm _

Kẻ \(\(DE\perp AC\)\)

Có \(\(\widehat{AOB}=\widehat{DOC}\)\)

Xét tam giác vuông \(\(DKO\)\), ta có :

\(\(AK=DO.\sin\widehat{DOK}\)\)hay \(\(AK=DO.\sin\widehat{AOB}\)\)

Do đó:

\(\(S_{\Delta ADC}=\frac{1}{2}.AC.DO.\sin\widehat{AOB}\left(1\right)\)\)

Tương tự :

\(\(S_{\Delta ACB}=\frac{1}{2}.AC.BO.\sin\widehat{AOB}\left(2\right)\)\)

Từ \(\(\left(1\right)\&\left(2\right)\Rightarrow S_{ABCD}=S_{\Delta ADC}+S_{\Delta ACB}=\frac{1}{2}.AC.\left(DO+BO\right).\sin\widehat{AOB}\)\)

\(\(\Leftrightarrow S_{ABCD}=\frac{1}{2}AC.BD.\sin\widehat{AOB}\left(dpcm\right)\)\)

_Minh ngụy_

Answer 2

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Cách 2 :

Ta có : \(\(\sin\widehat{AOD}=\sin\widehat{AOB}=\sin\widehat{COB}=\sin\widehat{COD}\left(=\sin a\right)\)\)

Mặt khác

\(\(2S_{\Delta AOD}=AO.OD.\sin a\)\)

\(\(2S_{AOB}=AO.OB.\sin a\)\)

\(\(2S_{BOC}=BO.OC.\sin a\)\)

\(\(2S_{COD}=DO.OC.\sin a\)\)

\(\(\Rightarrow2\left(S_{AOD}+S_{AOB}+S_{BOC}+S_{COD}\right)\)\)

\(\(=AO.OD.\sin a+AO.OB.\sin a+BO.OC.\sin a+DO.OC.\sin a\)\)

\(\(=\sin a.[\left(AO\left(OD+OB\right)+OC\left(OB+OD\right)\right)]\)\)

\(\(=\sin a.\left(OD+OB\right)\left(AO+OC\right)\)\)

\(\(=\sin a.BD.AC\)\)

\(\(\Rightarrow S_{\Delta AOD}+S_{\Delta AOB}+S_{\Delta BOC}+S_{\Delta COD}=\frac{1}{2}.AC.BD.\sin a\)\)

hay \(\(S_{ABCD}=\frac{1}{2}AC.BD.\sin a\)\)mà \(\(\sin\widehat{AOB}=\sin a\)\)

\(\(\Rightarrow S_{ABCD}=\frac{1}{2}AC.BD.\sin\widehat{AOB}\)\)

_Minh ngụy_