a) Xét hai tam giác vuông ABH và CAH có:$\widehat{ABH}$ = $\widehat{HAC}$ ( cùng phụ $\widehat{BAH}$$\rightarrow \Delta ABH \sim \Delta CAH$$\rightarrow \frac{AH}{CH}$ = $\frac{BH}{AH}$$\rightarrow$ AH2 = HB.HCABH^=HAC^" role="presentation" style="display: inline-block; line-height: normal; font-size: 18px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; color: rgb(51, 51, 51); font-family: 'times new roman'; position: relative;">ABH^=HA (cùng phụ với góc BAHBAH^" role="presentation" style="display: inline-block; line-height: normal; font-size: 18px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; color: rgb(51, 51, 51); font-family: 'times new roman'; position: relative;">BAH^Do đó, ΔABH∼ΔCAH" role="presentation" style="display: inline-block; line-height: normal; font-size: 18px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; color: rgb(51, 51, 51); font-family: 'times new roman'; position: relative;">ΔABH∼ΔCAH⇒AH2=BH.CH" role="presentation" style="display: inline-block; line-height: normal; font-size: 18px; word-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; color: rgb(51, 51, 51); font-family: 'times new roman'; position: relative;">
a) Xét hai tam giác vuông ABH và CAH có:$\widehat{ABH}$ = $\widehat{HAC}$ ( cùng phụ $\widehat{BAH}$$\rightarrow \Delta ABH \sim \Delta CAH$$\rightarrow \frac{AH}{CH}$ = $\frac{BH}{AH}$$\rightarrow$ AH2 = HB.HC