[hinh 8]tam giac dong dang

C

chonhoi110

Dễ dàng chứng minh $\Delta BHC \sim \Delta PBC$

$\rightarrow \dfrac{BH}{BP}= \dfrac{HC}{BC}\rightarrow \dfrac{BH}{BQ}=\dfrac{HC}{DC}$ (1)

Lại có $\widehat{HBC}=\widehat{HCD}= 90^0-\widehat{HCB}$ (2)

Từ (1) và (2) $\rightarrow \Delta HBQ\sim \Delta HCD (c.g.c)\rightarrow \widehat{BHQ}=\widehat{CHD}\rightarrow \widehat{DHQ}=\widehat{BHC}=90^0 \rightarrow đpcm$

Nguồn: VMF :D
 
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