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.