Câu 1:

\(ABCI\) là hình vuông \(\Rightarrow\left\{{}\begin{matrix}CD=\sqrt{IC^2+ID^2}=a\sqrt{2}\\AC=\sqrt{AB^2+BC^2}=a\sqrt{2}\end{matrix}\right.\)

\(\Rightarrow AC^2+CD^2=AD^2\Rightarrow\Delta ACD\) vuông cân tạiC

\(\Rightarrow OC\perp CD\) \(\Rightarrow CD\perp\left(SOC\right)\)

Từ O kẻ \(OH\perp SC\Rightarrow OH\perp\left(SCD\right)\) \(\Rightarrow OH\perp SD\)

\(\left\{{}\begin{matrix}BI\perp SO\\BI\perp OC\end{matrix}\right.\) \(\Rightarrow BI\perp\left(SOC\right)\Rightarrow BI\perp OH\)

\(SC=\sqrt{SO^2+OC^2}=a\sqrt{2}\) \(\Rightarrow SH=\frac{SO^2}{SC}=\frac{3a\sqrt{2}}{4}\)

Qua H kẻ đường thẳng song song CD cắt SD tại K

\(\frac{SH}{SC}=\frac{HK}{CD}\Rightarrow HK=\frac{SH.CD}{SC}=\frac{3a}{4}\)

Trên toa OI lấy điểm P sao cho \(OP=\frac{3a}{4}\)

\(\Rightarrow OHKP\) là hình chữ nhật \(\Rightarrow OH//KP\Rightarrow KP\) là đoạn vuông góc chung của \(BI\) và SD

\(\frac{1}{OH^2}=\frac{1}{SO^2}+\frac{1}{OC^2}\Rightarrow KP=OH=\frac{SO.OC}{\sqrt{SO^2+OC^2}}=\frac{a\sqrt{6}}{4}\)

Câu 2:

a/ Kẻ \(MH\perp AC\Rightarrow MH\perp\left(SAC\right)\)

\(\Rightarrow\widehat{MSH}\) là góc giữa SM và (SAC)

\(SM=\sqrt{SA^2+\left(\frac{AB}{2}\right)^2}=a\sqrt{10}\) ; \(MH=\frac{1}{2}\frac{2a\sqrt{3}}{2}=\frac{a\sqrt{3}}{2}\)

\(sin\widehat{MSH}=\frac{MH}{SM}=\frac{\sqrt{30}}{20}\Rightarrow\widehat{MSH}\approx15^053'\)

b/ \(\left\{{}\begin{matrix}MC\perp AB\\MC\perp SA\end{matrix}\right.\) \(\Rightarrow MC\perp\left(SAB\right)\)

\(\Rightarrow\widehat{SMA}\) là góc giữa \(\left(SMC\right)\) và \(\left(ABC\right)\)

\(tan\widehat{SMA}=\frac{SA}{AM}=3\Rightarrow\widehat{SMA}\approx71^033'\)

c/ Gọi N là trung điểm AC \(\Rightarrow NG=\frac{1}{3}NS\) (t/c trọng tâm)

\(\Rightarrow d\left(G;\left(SAB\right)\right)=\frac{1}{3}d\left(N;\left(SAB\right)\right)\)

Từ N kẻ \(NK\perp AB\Rightarrow NK\perp\left(SAB\right)\)

\(\Rightarrow NK=d\left(N;\left(SAB\right)\right)\)

\(NK=\frac{1}{2}.\frac{2a\sqrt{3}}{2}=\frac{a\sqrt{3}}{2}\Rightarrow d\left(G;\left(SAB\right)\right)=\frac{a\sqrt{3}}{6}\)

Bài 2:

d/ Do \(AM=BM\Rightarrow d\left(B;\left(SMC\right)\right)=d\left(A;SMC\right)\)

Theo cmt ta có \(CM\perp\left(SAB\right)\)

Từ A kẻ \(AP\perp SM\Rightarrow AP\perp\left(SMC\right)\)

\(\Rightarrow AP=d\left(A;\left(SMC\right)\right)=d\left(B;\left(SMC\right)\right)\)

\(\frac{1}{AP^2}=\frac{1}{SA^2}+\frac{1}{AM^2}\Rightarrow AP=\frac{SA.AM}{\sqrt{SA^2+AM^2}}=\frac{3a\sqrt{10}}{10}\)

e/

Do \(MN//BC\) (t/c đường trung bình) \(\Rightarrow BC//\left(SMN\right)\)

\(\Rightarrow d\left(BC;SM\right)=d\left(BC;\left(SMN\right)\right)=d\left(B;\left(SMN\right)\right)\)

Mà \(AM=BM\Rightarrow d\left(B;\left(SMN\right)\right)=d\left(A;\left(SMN\right)\right)\)

Từ A kẻ \(AQ\perp MN\Rightarrow MN\perp\left(SAQ\right)\)

Từ A kẻ \(AT\perp SQ\Rightarrow AT\perp\left(SMN\right)\)

\(\Rightarrow AT=d\left(A;\left(SMN\right)\right)=d\left(BC;SM\right)\)

\(AQ=\frac{1}{2}\frac{2a\sqrt{3}}{2}=\frac{a\sqrt{3}}{2}\)

\(\frac{1}{AT^2}=\frac{1}{AQ^2}+\frac{1}{SA^2}\Rightarrow AT=\frac{SA.AQ}{\sqrt{SA^2+AQ^2}}=\frac{3a\sqrt{13}}{13}\)

a) Ta có: \(\widehat{B}=120^o,\widehat{A}=90^o\Rightarrow\widehat{C}+\widehat{D}=360^o-\widehat{A}-\widehat{B}=150^o\)

CO, DO là hai tia phân giác góc C và góc D

=> \(\widehat{C_1}+\widehat{D_1}=\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}=\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)=\frac{1}{2}.150^o=75^o\)

=> \(\widehat{COD}=180^o-\left(\widehat{C_1}+\widehat{D_1}\right)=180^o-75^o=105^o\)

b) 

Xét tam giác COD

Ta có: \(\widehat{COD}=180^o-\left(\widehat{C_1}+\widehat{D_1}\right)=180^o-\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)\)

Vì: \(\widehat{C_1}+\widehat{D_1}=\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}=\frac{1}{2}\left(\widehat{C}+\widehat{D}\right)\)

Mặt khác: Xét tứ giác ABCD ta có: \(\widehat{C}+\widehat{D}=360^o-\widehat{A}-\widehat{B}\)

=> \(\widehat{COD}=180^o-\frac{1}{2}\left(360^o-\widehat{A}-\widehat{B}\right)=\frac{1}{2}\widehat{A}+\frac{1}{2}\widehat{B}\)

c) Tương tự ta cũng chứng minh dc:

\(\widehat{BIA}=\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}\)

=> \(\widehat{COD}+\widehat{BIA}=\frac{1}{2}\widehat{A}+\frac{1}{2}\widehat{B}+\frac{1}{2}\widehat{C}+\frac{1}{2}\widehat{D}=\frac{1}{2}\left(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}\right)=\frac{1}{2}.360^o=180^o\)

=>\(\widehat{FOE}+\widehat{EIF}=180^o\)

=> \(\widehat{OEI}+\widehat{IFO}=180^o\)

Vậy tứ giác EIF có các góc đối bù nhau!

Ta có BAD + ABC + BCD + CDA = 360 độ

ADC + BCD = 360 - 120 - 90 = 150 độ

=> BCO = OCD = 1/2 BCD

=> ADO = ODC = 1/2 ADC

=> ODC + OCD = 1/2 ODC + 1/2 OCD = ODC+OCD/2

=> ODC + OCD = 150 /2 =75 độ

Mà ODC + OCD +DOC = 180 độ

=> DOC = 180 - 75 = 105 độ

B) COD = 180 - (ODC + OCD) 

=> COD = 180 - 1/2ADC + 1/2 BCD

Mà ADC + BCD = 360 - ( BAD + ABC)

COD = 180 - [ 360 - 1/2(BAD + ABC )]

Tách ra đi bạn

ai lam thi lam di 

đc có tí điểm bắt lm 5 câu dài ko ai muốn lm