P65' s 8: Angle DBC= Angle ADB= Angle BDC Obviously, CBD is an isosceles triangle BC=CD=AB, so BC=Y, AD=X, so X+2Y=20, and X+Y= 12, so X=4, that is, the upper bottom is 4cm.
The rectangular proof when the point 10:O of P66 is the AC midpoint;
∫EC is the bisector of ∞∠BCA.
∴∠eco =∞∠ECB and FC are bisectors of the outer corner of △ABC.
∴∠OCF= 1/2( 180-∠BCA)
∴∠ECO+∠OCF=90
∴∠ECF = 90° parallel to MN BC
∴∠OEC=∠ECB=∠ECO
∴OE=OC =OC in the same way.
∴OE=OF
∴O is the midpoint of EF and O is the midpoint of AC. EF AC is the diagonal of the quadrilateral AECF.
∴AECF is a parallelogram, and∠ ∠ECF = 90°.
∴AECF is a rectangle.