Cho anh hay nhất nhé

Giải thích các bước giải:

a.Xét $\Delta APE,\Delta APH$ có:

Chung $AP$

$\widehat{APE}=\widehat{APH}(=90^o)$

$PE=PH$

$\to \Delta APE=\Delta APH(c.g.c)$

Xét $\Delta AQH,\Delta AQF$ có:

Chung $AQ$

$\widehat{AQH}=\widehat{AQF}(=90^o)$

$QH=QF$

$\to \Delta AQH=\Delta AQF(c.g.c)$ 

b.Từ câu a $\to AE=AH, \widehat{EAP}=\widehat{PAH}$

                            $AH=AF, \widehat{HAQ}=\widehat{AQF}$

$\to AE=AF(=AH)$

Mặt khác:

$\widehat{EAF}=\widehat{EAP}+\widehat{PAH}+\widehat{HAQ}+\widehat{QAF}=\widehat{PAH}+\widehat{PAH}+\widehat{HAQ}+\widehat{HAQ}=2\widehat{PAH}+2\widehat{HAQ}=2(\widehat{PAH}+\widehat{HAQ})=2\widehat{PAQ}=2\widehat{BAC}=180^o$

$\to E, A, F$ thẳng hàng

Mà $AE=AF\to A$ là trung điểm $EF$

c.Xét $\Delta ABE,\Delta ABH$ có:

Chung $AB$

$\widehat{BAE}=\widehat{PAE}=\widehat{PAH}=\widehat{BAH}$

$AE=AH$

$\to \Delta ABE=\Delta ABH(c.g.c)$

$\to \widehat{BEA}=\widehat{BHA}=90^o$

$\to BE\perp AE\to BE\perp EF$

Tương tự chứng minh được $CF\perp EF$

$\to BE//CF$