Question

如图,四边形ABcD为菱形,ACBD是对角线,∠ABC=30°求证AB²=AC*BD

Answer 1

证明：设AC与BD相交于O点

在直角三角形AOB中

∠AOB=1/2∠ABC=1/2*30°=15°

从而 OA=AB*sin∠AOB,OB=AB*cos∠AOB

∵sin∠AOB=sin15°

=sin30°/2

=√((1-cos30°)/2)

=√((1-√3/2)/2)

=√(2-√3)/2

cos∠AOB=cos15°

=cos30°/2

=√((1+cos30°)/2)

=√((1+√3/2)/2)

= √(2+√3)/2

∴OA*OB=AB*√(2-√3)/2*AB*√(2+√3)/2

=AB^2/4

又 OA=1/2AC,OB=1/2BD

∴AB^2=4*OA*OB

=4*1/2AC*1/2BD

=AC*BD

则 AB²=AC*BD.