lần sau bạn nên làm 3 post vì thường ngta chỉ muốn làm 1 câu

\(5,\\ a,\left\{{}\begin{matrix}AB=CD\left(gt\right)\\AD=BC\left(gt\right)\\AC.chung\end{matrix}\right.\Rightarrow\Delta ABC=\Delta CDA\left(c.c.c\right)\\ b,\Delta ABC=\Delta CDA\left(cm.trên\right)\\ \Rightarrow\left\{{}\begin{matrix}\widehat{CAB}=\widehat{DCA}\\\widehat{CAD}=\widehat{ACB}\end{matrix}\right.\left(các.cặp.góc.tương.ứng\right)\)

Mà các cặp góc này ở vị trí so le trong nên \(AB//CD;AD//BC\)

what the hell?

What 

a) \(\widehat{AED}=\widehat{AFD}=90^o\) nên \(E,F\) cùng nhìn \(AD\) dưới góc vuông suy ra \(AEDF\) nội tiếp. 

suy ra \(\widehat{AEF}=\widehat{ADF}\).

mà \(\widehat{ADF}=\widehat{ACD}\) (vì cùng phụ với góc \(\widehat{DAC}\))

suy ra \(\widehat{AEF}=\widehat{ACD}\Rightarrow\widehat{BEF}+\widehat{FCB}=180^o\) suy ra \(BEFC\) nội tiếp.

b) \(\Delta GBE\sim\Delta GFC\left(g.g\right)\)

suy ra \(GB.GC=GE.GF\).

\(\Delta GDE\sim\Delta GFD\left(g.g\right)\)

suy ra  \(GD^2=GE.GF\).

\(ACBH\) nội tiếp suy ra \(GB.GC=GH.GA\)

suy ra \(GD^2=GH.GA\)

\(\Rightarrow\Delta GHD\sim\Delta GDA\left(c.g.c\right)\)

\(\Rightarrow\widehat{GHD}=\widehat{GDA}=90^o\)

suy ra \(DH\) vuông góc với \(AG\). 

a: Xét ΔODB vuông tại D vàΔOCA vuông tại C có

OB=OA

góc O chung

=>ΔODB=ΔOCA

b: ΔODB=ΔOCA

=>góc OBD=góc OAC

góc OBD+góc IBA=góc OBA

góc OAC+góc IAB=góc OAB

mà góc OBD=góc OAC và góc OAB=góc OBA

nên góc IAB=góc IBA

=>ΔIAB cân tại I

c: IC=ID

ID<IA

=>IC<IA

1 Where did you go?

2 Who did you go with?

3 How did you get there?

4 What did you do during the day?

5 Did you have a good time?

1. Where did you go?

Where was you going?

2. Who did you go with?

Who was you going with?

3.  How did you get there?

How was you getting there?

5. What did you during the day?

What was you doing during the day?

6. Did you have a good time?

Was you having a good time?

Câu 5: 

\(x=\dfrac{6^2}{10}=3.6\left(cm\right)\)

y=10-3,6=6,4(cm)