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Gọi giao điểm của hai đường chéo là O giao điểm của hai cạnh bên là S,giao điểm của SO với AB,CD lần lượt là X,Y.
Ta có AX//YC nên theo định lý Ta lét ta có:
\(\frac{AX}{YC}\)=\(\frac{AO}{OC}\)=\(\frac{AB}{DC}\)=\(\frac{AX}{DY}\)
=>YC=DY
Vậy Y là trung điểm của DC.
Ta có AB//DC theo định lý Ta-lét ta có:
\(\frac{AX}{DY}\)=\(\frac{SX}{XY}\)=\(\frac{XB}{YC}\)
mà DY=YC(c/m trên)
=>AX=XB=>X là trung điểm của AB
Vậy giao điểm của SO với AB,CD tại trung điểm của các cạnh đó
=>đpcm
Ta cũng dễ dàng chứng mình được đường thẳng chứa 4 điểm đó là trùng trực của hai cạnh đấy sao khi chừng minh chúng thẳng hàng ở trên nhé!
Gọi giao điểm của hai đường chéo là O giao điểm của hai cạnh bên là S,giao điểm của SO với AB,CD lần lượt là X,Y.
Ta có AX//YC nên theo định lý Ta lét ta có:
AXYCAXYC=AOOCAOOC=ABDCABDC=AXDYAXDY
=>YC=DY
Vậy Y là trung điểm của DC.
Ta có AB//DC theo định lý Ta-lét ta có:
AXDYAXDY=SXXYSXXY=XBYCXBYC
mà DY=YC(c/m trên)
=>AX=XB=>X là trung điểm của AB
Vậy giao điểm của SO với AB,CD tại trung điểm của các cạnh đó
=>đpcm
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
a: Trên tia đối của tia MA, lấy I sao cho MA=MI. Gọi K là giao điểm của AM và BC
Xét tứ giác EAHI có
M là trung điểm chung của EH và AI
=>EAHI là hình bình hành
=>AH=IE
mà AH=AC(ACGH là hình vuông)
nên IE=AC
Ta có: EAHI là hình bình hành
=>\(\hat{EAH}+\hat{AEI}=180^0\) (1)
Ta có: \(\hat{EAH}+\hat{EAB}+\hat{CAH}+\hat{BAC}=360^0\)
=>\(\hat{EAH}+\hat{BAC}=360^0-90^0-90^0=180^0\) (2)
Từ (1),(2) suy ra \(\hat{AEI}=\hat{BAC}\)
Xét ΔAEI và ΔBAC có
AE=BA
\(\hat{AEI}=\hat{BAC}\)
EI=AC
Do đó: ΔAEI=ΔBAC
=>\(\hat{EAI}=\hat{ABC}\)
Ta có: \(\hat{EAI}+\hat{EAB}+\hat{BAK}=180^0\)
=>\(\hat{EAI}+\hat{BAK}=180^0-90^0=90^0\)
=>\(\hat{BAK}+\hat{ABK}=90^0\)
=>ΔAKB vuông tại K
=>AK⊥BC tại K
=>AM⊥BC
bạn ghi có sai đề ko chứ ? 1 tam giác làm sao mà có 1 góc =180 độ đc?