Let BC = a AC = b ab = C.

The vertical feet from o to BC AC AB are D E F respectively.

DO * BC = 1/3 *(AC * BC)= 1/3ab OD = 1/3b CE = OD AE = 2/3b

Similarly, OE =1/3acd = oebd =1/2ao2 = ae2+oe2 = 4/9b2+a2/9.

ob^2=bd^2+oe^2 = 4/9 a^2 +b^2/9 oc^2=cd^2+od^2 =a^2/9 +b^2/9

∴ the square of OA+the square of OB = the square of 5oc.

This is a classic question, and the answer is available online. )

2. Solution:

∵PA^2+PC^2=PB^2+PD^2

(ps: Proof of the above formula.

Make EF//AD by crossing at point p, AB at point e and CD at point F.

Do GH//AB through P, G through AD, and H through BC.

Because rectangular ABCD

So angle AEP= angle PFD=90 degrees, GP=AE=DF, PH=BE=FC.

From Pythagorean Theorem:

PA^2=PG^2+PE^2

PB^2=PH^2+PE^2

PC^2=PF^2+PH^2

PD^2=PF^2+PG^2

)

∴PD=√(PA^2+PC^2-PB^2)

=√(5^2+ 14^2- 10^2)

= 1 1

This question is so simple that people are speechless. )

3. Solution:

Turn △ APB 90 to △BCE clockwise around point B, and connect PE to make ∠ PBE = 90.

∠∠APB =∠BEC，BE=BP=2，CE=AP= 1

∴△PBE is an isosceles right triangle, ∠ PEB = 45.

According to Pythagorean theorem, PE 2 = Pb+Be 2 = 2 2+2 2 = 8.

PE^2+CE^2=8+ 1=9=3^2=PC^2

∴△PEC is Rt△, and∠ PEC = 90.

∴∠apb=∠bec=∠peb+∠pec=45+90 = 135

(ps: another classic)

4. There is something wrong with the topic. 2MN is not equal to EC.

forget