<Qua-2026-815>, hint

AC and BD intersect at Q.

AQ+DQ>AD , BQ+CQ>BC---> AC+BD>BC+AD, similarly AC+BD>AB+CD

Hence AC+BD>1/2(AB+BC+CD+DA)

AB+BC>AC, AD+CD>AC  AB+BC+CD+DA>2AC. Similarly AB+BC+CD+DA>2BD

Hence AC+BD<AB+BC+CD+DA

<Qua-2026-815>, general, perimeter

<South Africa MO 2001>

 

<Qua-2026-814 >, hint

<Qua-2026-814>, square, perimter

<South Africa MO 2000>

 

<Qua-2026-813>, hint

Let <BAD=2x, then <BAK=<BKA=<CKL=<CLK=x 

Let O be the circumcenter of triangle CKL, OK=OC (radius), <KOC=2KLC=2x.

BK=BA=CD, <OCD=<BCD+<OCK=2x+90-x=90+x, <OKB=180-<OKC(90-x)=90+x

Hence triangles OCD and OKB are congruent and <OBC=<ODC and O,B,C,D are cyclic.