Question

等腰梯形ABCD,AB‖CD,求证:AC²=AD²+AB·CD

Answer 1

过C作AB的垂线于E,BE=(AB-CD)/2

由余弦定理:

AC^2=AD^2+CD^2-2AD*CD*cos∠ADC ①

AC^2=BC^2+AB^2-2BC*AB*cos∠ABC ②

而BC=AD,∠ADC=180°-∠ABC,cos∠ADC=-cos∠ABC

(①+②)/2得AC^2=AD^2+(AB^2+CD^2)/2-BC(AB-CD)cos∠ABC

=AD^2+(AB^2+CD^2)/2-(AB-CD)*BE

=AD^2+(AB^2+CD^2)/2-(AB-CD)^2 /2

=AD^2+AB*CD