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