(Câu 6 chưa ra)

1. 

Theo Pytago: $BC= \sqrt{AB^2+AC^2}= 10cm$

$\Delta$ AHB và $\Delta$ CAB có $\widehat{ABH}$ chung, $\widehat{AHB}= \widehat{CAB}= 90^o$ 

=> $\Delta$ AHB $\backsim$ $\Delta$ CAB (g.g)

=> $\frac{AH}{AB}= \frac{AC}{BC}$

$\Leftrightarrow AH= 4,8cm$

2.

$\Delta$ AEH $\backsim$ $\Delta$ AHB (g.g) vì $\widehat{EAH}$ chung, $\widehat{AEH}= \widehat{AHB}= 90^o$

3.

$\Delta$ AFH $\backsim$ $\Delta$ AHC (g.g) vì $\widehat{ACH}$ chung, $\widehat{AFH}= \widehat{AHC}= 90^o$

=> $\frac{AH}{AF}= \frac{AC}{AH}$

$\Leftrightarrow AH^2= AC.AF$

4.

$\Delta$ AHB $\backsim$ $\Delta$ CAB (g.g) (theo a) 

Mà $\Delta$ AEH $\backsim$ $\Delta$ AHB (g.g) (theo b)

=> $\Delta$ AEH $\backsim$ $\Delta$ CAB 

$\Delta$ AEH = $\Delta$ AFE (c.g.c, theo hình chữ nhật AEHF) 

=> $\Delta$ AFE $\backsim$ $\Delta$ CAB

5.

$AF= \frac{AH^2}{AC}= 2,88cm$

CMTT, $AE= \frac{AH^2}{AB}= 3,84cm$

=> $S_{AEF}= 0,5.2,88.3,84= 5,5296cm^2$

$S_{ABC}= 0,5AB.AC= 0,5.6.8= 24cm^2$

=> $S_{BEFC}= 24-5,5296= 18,4704cm^2$