Question

线段BD=1/3AB=1/4CD,点M,N分别是线段AB,CD的中点,且MN=20cm,求AC的长

Answer 1

1/3AB=1/4CD CD=4/3AB 1/2CD=2/3AB DN=1/2CD

BN=1/2CD-BD=2/3AB-1/3AB=1/3AB

MN=MD+BD+BN

MN=(1/2AB-1/3AB)+1/3AB+(DN-DB）

=1/6AB+1/3AB+2/3AB-1/3AB

=5/6AB

AB=20X6/5=24

CD=4/3AB=4/3X24=32 DB=1/3AB=1/3X24=8 AC=AB+CD-BD=24+32-8=48