Answers

2016-03-20T02:27:35+05:30
Let abcd be a square
join po
ao = po given
therefore. angle OAP = angle opa opp. angles of equal side

but diagonals of square are equal and bisect each other
AO = bo
ao=po. given
bo=po



angle opb = oba
now, opa + apo + poa = 180
x+x+poa=180
poa =180-2x. eq.1

opb + pbo+ bop =180
bop=180-2x. eq.2


from eq. 1 and 2
poa=bop
3poa=3bop
1 5 1
  • Brainly User
2016-03-20T05:50:44+05:30
ABCD is a square and AC, BD the 2 diagonals intersect each other at O. If P is a point on AB such that AO=OP, prove that 3of anglePOB=3of angleAOP. 



Let ∠POB = x°
It is known that diagonals of a square bisect the angles.

∴ ∠OBP = ∠OAP = (90°/2) = 45°

Using exterior angle property for ∆OPB, ∠OPA = ∠OBP + ∠POB = 45° + x° In OAP, OA = AP .



∴ ∠OPA = ∠AOP ⇒ ∠AOP = 45° + x° …

(1) It is known that diagonals of square are perpendicular to each other.
 

∴ ∠AOP + ∠POB = 90° ⇒ 45° + x° + x° = 90°

[Using (1)]

⇒ 2x = 90 – 45 = 45 ⇒ x = 22.5

∴ ∠POB = 22.5° and ∠AOP = 45° + 22.5° = 67.5° = 3 × 22.5°
 
⇒ ∠AOP = 3∠POB  
1 5 1
we have to prove that 3 of aop = 3 of pob
you done wrong