[Toán 8]Bài tập hình tuần

N

nhuquynhdat

Kẻ $AH \perp CD$

Ta có: $S_{ACD}= \dfrac{1}{2}.AH.CD$

$S_{BCD}= \dfrac{1}{2}.AH.CD$

$\to S_{ACD}=S_{BCD}$

$\to S_{AOD}+ S_{DOC}= S_{DOC}+S_{BOC} \to S_{AOD}= S_{BOC}$
 
H

huynhbachkhoa23

Ta có:
vì $AB//CD$
$\to S_{ACD}=S_{BDC}$
$\to S_{ACD}-S_{COD}=S_{BDC}-S_{COD}$
$\to S_{AOD}=S_{DOC}$
 
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