P64' s 5: extends the intersection of CE and AB at point G, because angle 1= angle 2, AE is the public side, angle AEC= angle AEG=90 degrees, so triangle AEC is congruent with triangle AEG, so EC=EG and angle 1= angle BCG AB=BC are at right angles to angle CBG, so triangle ABF and triangle.

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.