Because p is the midpoint between BC and two right triangles.

So AP=DP QP=QP.

So AQ=QD

15 extends GP to BC at o point.

Be educated

Angle EBD= angle EDB

Because right-angle EDB= right-angle DBC

PF is perpendicular to be and po is perpendicular to BC.

PF=PO

ABOG is rectangular.

PF+PG=PO+PG=AB

14.( 1) First of all, DB=AD is parallel to CF DE=CE.

The available triangle ADE is equal to triangle FCE.

So AD=CF

So DB=CF

(2) If the rectangle is (1), BD is parallel and equal to CF; If AC=BC, the angle BDC is equal to 90; If BF is connected, the quadrilateral is a rectangle.

16

1

Proof: ∫MN//BC

∴∠OEC=∠BCE

∴∠OFC=∠FCG

∠∠BCE =∠OCE(OE is the bisector of the inner corner of∠∠ ∠BCA)

∴∠OEC=∠OCE

∴OE=OC

∠∠OCF =∠FCG(OF is the bisector of ∠BCA).

∴∠OCF=∠OFC

∴OF=OC

∴OE=OF

2

O When moving to the midpoint of the AC side, the quadrilateral AECF is a rectangle. Prove:

∫OE = OCOE = OF

When o is the midpoint of communication, OA=OC.

∴OE=OC=OF=OA

∴ Quadrilateral AECF is a rectangle.