Kaya Renger Coordinator
06/09/2017 at 21:17-
Applying the Pythagorean theorem ,we have :
\(BC^2=AB^2+AC^2\)
\(\Leftrightarrow\left(HB+HC\right)^2=AH^2+HB^2+AH^2+HC^2\)
\(\Leftrightarrow HB^2+2HB.HC+HC^2=2AH^2+HB^2+HC^2\)
\(\Leftrightarrow2HB.HC=2AH^2\Leftrightarrow AH^2=HB.HC\)
Then : \(BH.BC=BH\left(BH+HC\right)=BH^2+BH.HC=BH^2+AH^2=AB^2\)
P/S : I think it's unnecessary to draw HD and HE
Kaya Renger selected this answer. -
hochiminh 07/09/2017 at 19:45
Áp dụng định lý Pythagore, chúng ta có:
B C 2 = A B 2 + A C 2BC2=AB2+AC2
⇔ ( H B + H C ) 2 = A H 2 + H B 2 + A H 2 + H C 2⇔(HB+HC)2=AH2+HB2+AH2+HC2
⇔ H B 2 + 2 H B . H C + H C 2 = 2 A H 2 + H B 2 + H C 2⇔HB2+2HB.HC+HC2=2AH2+HB2+HC2
⇔ 2 H B . H C = 2 A H 2 ⇔ Một H 2 = H B . H C⇔2HB.HC=2AH2⇔AH2=HB.HC
Sau đó: B H . B C = B H ( B H + H C ) = B H 2 + B H . H C = B H 2 + A H 2 = A B 2