6. it is proved that the intersection point d is DM, the vertical AB is in M DN, and the vertical AC is in n.

So the S triangle DFB= 1/2BF*DM.

S triangle DEC= 1/2*CE*DN

Angle AMD= sum of angles =90 degrees

So triangle AMD AND triangle AND are right triangles.

Because the areas of triangle DFB and triangle DEC are equal.

BF=CE

So DM=DN

Because AD=AD

So right triangle AMD AND right triangle AND are congruent (HL)

So angle MAD= angle NAD

So AD bisects the angle BAC

7. It is proved that the intersection f makes FN perpendicular to AP and n..

So ANF angle = PNF angle =90 degrees.

Because the quadrilateral ABCD is a square

So AB=BC=DC=AD

Angle B= angle C= angle D=90 degrees.

So angle B= angle ANF=90 degrees.

So triangle ABF and triangle ANF are right triangles.

Because AF divides BAP equally

So angle BAF=\ angle PAF

Because AF=AF

So the right triangle ABF and the right triangle ANF are congruent (HL)

So AB=AN

BF=NF

Because AP=AN+NP

AP=AB+CP

So NP=CP

Angle PNF= angle C=90 degrees (confirmed)

So triangle PNF and triangle PCF are right triangles.

Because PF=PF

So the right triangle PNF and the right triangle PCF are congruent (HL)

So NF=CF

So BF=CF= 1/2BC.

Because point m is the midpoint of the CD.

So DM= 1/2DC.

So BF=DM

Angle B= Angle D=90 degrees (confirmed)

AB=AD (authentication)

So triangle ABF and triangle ADM are congruent (SAS)

So angle BAF= angle DAM