Question

Plz answer these to questions to earn 15 pts…

Q1 :
OP bisects Angle AOC, OQ bisects
Angle BOC, OP is perpendicular to OQ. Prove that A,O,B are collinear …

Q2 :
In a quadrilateral ABCD diagonals intersect at O. Prove that AB+BC+CD+AD < 2(AC+BD)…

Q3 :
"There exist at least 3 points which aren't in the same straight line."
Which of Euclid's postulates is followed in this statement? Explain…

Note :
Pics of written solutions are recommended as there are 3 questions …

Answer 1

1) given: OP bisects ∠AOC and OQ bisects ∠BOC.

TPT: AOB are collinear points.

proof:

OP bisects ∠ AOC ⇒ ∠AOP = ∠ POC = 1/2 ∠AOC

OQ bisects ∠BOC ⇒ ∠BOQ = ∠ COQ = 1/2 ∠BOC

again , given that

OP ⊥ OQ

∠ POC + ∠COQ = 90 deg

therefore

1/2∠AOC + 1/2∠BOC = 90

1/2(∠AOC + ∠BOC) = 90

∠AOC + ∠BOC = 90 * 2

∠AOC + ∠BOC = 180

∠ AOB = 180 deg

thus AOB is a straight line and A, O and B are collinear points.

2) Hope you have the diagram with you. 

Now by the property of triangles, sum of any two sides > the third side. 
Therefore, AB + BC > AC Same way CD +DA > AC 
Similar way we can get DA + AB > BD and BC + CD > BD 
Now adding all these inequalities, 2(AB + BC +CD +DA) > 2(AC + BD) 
Dividing both sides by 2 we get, (AB + BC +CD +DA) > (AC + BD) 
Thus, the given first identity is proved. 
For the second identity, 
2AC > AB + BC Similar way, 2AC > CD + DA 
Also, 2BD > DA + AB and 2BD > BC + CD 
Adding the inequalities, we get 4(AC + BD) > 2(AB + BC +CD +DA) 
Dividing by 2 on both sides, 
we arrive at 2(AC + BD) > (AB + BC +CD +DA) 
or (AB + BC +CD +DA) < 2(AC + BD) Hence second identity is proved.

3)  states that for the given points A and B we can take point C not lying on the lines.These postulates do not follow from Euclid’s postulates.  However they follow from the 1st axiom of incidence geometry and 2nd axiom of neutral geometry: "there are exist at least 3 non-collinear points".

Answer 2

Answer:

1) given: OP bisects ∠AOC and OQ bisects ∠BOC.TPT: AOB are collinear points.

proof:

OP bisects ∠ AOC ⇒ ∠AOP = ∠ POC = 1/2 ∠AOC

OQ bisects ∠BOC ⇒ ∠BOQ = ∠ COQ = 1/2 ∠BOC

again , given that

OP ⊥ OQ

∠ POC + ∠COQ = 90 deg

therefore

1/2∠AOC + 1/2∠BOC = 90

1/2(∠AOC + ∠BOC) = 90

∠AOC + ∠BOC = 90 * 2

∠AOC + ∠BOC = 180

∠ AOB = 180 deg

thus AOB is a straight line and A, O and B are collinear points.

2) Hope you have the diagram with you. 

Now by the property of triangles, sum of any two sides > the third side. 

Therefore, AB + BC > AC Same way CD +DA > AC 

Similar way we can get DA + AB > BD and BC + CD > BD 

Now adding all these inequalities, 2(AB + BC +CD +DA) > 2(AC + BD) 

Dividing both sides by 2 we get, (AB + BC +CD +DA) > (AC + BD) 

Thus, the given first identity is proved. 

For the second identity, 

2AC > AB + BC Similar way, 2AC > CD + DA 

Also, 2BD > DA + AB and 2BD > BC + CD 

Adding the inequalities, we get 4(AC + BD) > 2(AB + BC +CD +DA) 

Dividing by 2 on both sides, 

we arrive at 2(AC + BD) > (AB + BC +CD +DA) 

or (AB + BC +CD +DA) < 2(AC + BD) Hence second identity is proved.

3)  states that for the given points A and B we can take point C not lying on the lines.These postulates do not follow from Euclid’s postulates.  However they follow from the 1st axiom of incidence geometry and 2nd axiom of neutral geometry: "there are exist at least 3 non-collinear points".