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乇尺尺のﾚ · Accepted Answer

a)a/ Xét ΔABC và ΔDEC ta có:$\widehat{C}$ $chung$$\widehat{ABC}=\widehat{DEC}\left(gt\right)$$\Rightarrow\Delta ABC$ ∼ $\Delta DEC\left(g.g\right)$b/ $Vì$ $\Delta ABC$ ∼$\Delta DEC\left(g.g\right)\left(cmt\right)$$\Rightarrow\dfrac{AB}{DE}=\dfrac{AC}{DC}$b)a/$Xét\Delta ABE$ $và$ $\Delta KBH$ $ta$ $có:$$\widehat{BEA}=\widehat{KHB}\left(gt\right)$$\widehat{HBK}=\widehat{EBA}\left(đđ\right)$$\Rightarrow\Delta ABE$∼$\Delta KBH\left(g.g\right)$b/ $Vì$ $\Delta ABE$∼$\Delta KBH\left(g.g\right)\left(cmt\right)$$\rightarrow\dfrac{AE}{KH}=\dfrac{AB}{KB}$$\Rightarrow AE.KB=KH.AB$c)a/$Xét$ $\Delta AHQ$ $và$ $\Delta FEQ$  $ta$ $có$:$\widehat{Q}$  $chung$$\widehat{HAQ}=\widehat{EFQ}=90^0$$\Rightarrow\Delta AHQ$∼$\Delta FEQ\left(g.g\right)$b/ $Vì$ $\Delta AHQ$∼$\Delta FEQ\left(g.g\right)\left(cmt\right)$$\rightarrow\dfrac{AH}{EF}=\dfrac{HQ}{EQ}$$\Rightarrow AH.EQ=EF.HQ$d)a/$Xét\Delta DEF$ $và$ $\Delta DAE$ $ta$ $có:$$\widehat{D}$  $chung$$\widehat{DEF}=\widehat{EAD}=90^0$$\Rightarrow\Delta DEF$∼$\Delta DAE\left(g.g\right)\left(1\right)$b/$Xét\Delta EAF$ $và$ $\Delta FED$ $ta$ $có$:$\widehat{F}$ $chung$$\widehat{EAF}=\widehat{DEF}=90^0$$\Rightarrow\Delta EAF$∼$\Delta FED\left(g.g\right)\left(2\right)$$Từ\left(1\right)và\left(2\right)suy$ $ra$ $\Delta DEF$∼$\Delta DAE$∼$\Delta AEF$$\Rightarrow\Delta DAE$∼$\Delta EAF$c/$Vì\Delta DAE$∼$\Delta EAF\left(cmt\right)$$\rightarrow\dfrac{EA}{DA}=\dfrac{AF}{EA}$$\rightarrow EA.EA=DA.AF$$\Rightarrow EA^2=DA.AF$ Câu trả lời: a)a/ Xét ΔABC và ΔDEC ta có:$\widehat{C}$ $chung$$\widehat{ABC}=\widehat{DEC}\left(gt\right)$$\Rightarrow\Delta ABC$ ∼ $\Delta DEC\left(g.g\right)$b/ $Vì$ $\Delta ABC$ ∼$\Delta DEC\left(g.g\right)\left(cmt\right)$$\Rightarrow\dfrac{AB}{DE}=\dfrac{AC}{DC}$b)a/$Xét\Delta ABE$ $và$ $\Delta KBH$ $ta$ $có:$$\widehat{BEA}=\widehat{KHB}\left(gt\right)$$\widehat{HBK}=\widehat{EBA}\left(đđ\right)$$\Rightarrow\Delta ABE$∼$\Delta KBH\left(g.g\right)$b/ $Vì$ $\Delta ABE$∼$\Delta KBH\left(g.g\right)\left(cmt\right)$$\rightarrow\dfrac{AE}{KH}=\dfrac{AB}{KB}$$\Rightarrow AE.KB=KH.AB$c)a/$Xét$ $\Delta AHQ$ $và$ $\Delta FEQ$  $ta$ $có$:$\widehat{Q}$  $chung$$\widehat{HAQ}=\widehat{EFQ}=90^0$$\Rightarrow\Delta AHQ$∼$\Delta FEQ\left(g.g\right)$b/ $Vì$ $\Delta AHQ$∼$\Delta FEQ\left(g.g\right)\left(cmt\right)$$\rightarrow\dfrac{AH}{EF}=\dfrac{HQ}{EQ}$$\Rightarrow AH.EQ=EF.HQ$d)a/$Xét\Delta DEF$ $và$ $\Delta DAE$ $ta$ $có:$$\widehat{D}$  $chung$$\widehat{DEF}=\widehat{EAD}=90^0$$\Rightarrow\Delta DEF$∼$\Delta DAE\left(g.g\right)\left(1\right)$b/$Xét\Delta EAF$ $và$ $\Delta FED$ $ta$ $có$:$\widehat{F}$ $chung$$\widehat{EAF}=\widehat{DEF}=90^0$$\Rightarrow\Delta EAF$∼$\Delta FED\left(g.g\right)\left(2\right)$$Từ\left(1\right)và\left(2\right)suy$ $ra$ $\Delta DEF$∼$\Delta DAE$∼$\Delta AEF$$\Rightarrow\Delta DAE$∼$\Delta EAF$c/$Vì\Delta DAE$∼$\Delta EAF\left(cmt\right)$$\rightarrow\dfrac{EA}{DA}=\dfrac{AF}{EA}$$\rightarrow EA.EA=DA.AF$$\Rightarrow EA^2=DA.AF$

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