Question

[Duyên Nguyễn Đặng Như]

Cho tam giac ABC co BC=15cm, AC=20cm, AB=25cm

a) Tinh do dai duong cao CH cua tam giac ABC

b) Goi CD la duong phan giac cua \(\Delta\)ACH. Cm \(\Delta\)BCD can

c) Cmr \(BC^2\)+\(CD^2\)+\(BD^2\)=\(3CH^2\)+\(2BH^2\)+\(DH^2\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: Xét ΔBCA có \(BA^2=CA^2+CB^2\)

nênΔBCA vuông tại C

\(CH=\dfrac{CA\cdot CB}{AB}=12\left(cm\right)\)

b: Ta có: \(\widehat{BCD}+\widehat{ACD}=90^0\)

\(\widehat{BDC}+\widehat{HCD}=90^0\)

mà \(\widehat{ACD}=\widehat{HCD}\)

nên \(\widehat{BCD}=\widehat{BDC}\)

hay ΔBCD cân tại B