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Agnishom

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From: Riemann Sphere

Registered: 2011-01-29

Posts: 24,996

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Geometry; Quadrilateral; Midpoints

A quadrilateral PQRS  has angular  bisectors of angle S and angle P which meet inside the quadrilateral at a point O.

Is angle Q + angle P = 2 x Angle SOP?

No other clues are visible

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reconsideryouranswer

Member

Registered: 2011-05-11

Posts: 171

Re: Geometry; Quadrilateral; Midpoints

A quadrilateral PQRS  has angular bisectors of angle S
and angle P which meet inside the quadrilateral at a point O.

Does the measure of angle Q + the measure of angle P =
twice the measure of angle SOP?

No other clues are visible

For instance, it *does* work for a quadrilateral when it is a square.

However, I am looking for a counterexample with the following:

Last edited by reconsideryouranswer (2011-10-15 15:43:47)

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Bob

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Registered: 2010-06-20

Posts: 10,583

Re: Geometry; Quadrilateral; Midpoints

hi Agnishom

I've set this up using Sketchpad and tried a variety of quadrilaterals.

The angle property is rarely true.  For a given set of fixed positions for 3 of the points I can vary the fourth until the property holds.

Why did you think it might?

Was this a given problem to prove?

Any other information such as (i) it's a cyclic quad,  or (ii) certain sides are parallel / equal ?

Bob 

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Agnishom

Real Member

From: Riemann Sphere

Registered: 2011-01-29

Posts: 24,996

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Re: Geometry; Quadrilateral; Midpoints

What if it is A parallelogram or trapezium?

Last edited by Agnishom (2011-10-15 21:55:31)

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'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'
'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'
I'm not crazy, my mother had me tested.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,583

Re: Geometry; Quadrilateral; Midpoints

hi Agnishom,

I tried making SR parallel to PQ.  See diagram below.

I found that angle SOP = 90, for all possible positions of PQRS.

That is easily proved:

Let angle RSP = 2x and angle SPQ = 2y.

Then, as PQ parallel SR => 2x + 2y = 180

=> x + y = 90

So in triangle SOP, the angle SOP = 90.

So, if you want SPQ + PQR = 2 x SOP you need

SPQ + PQR = 180       =>     PS parallel QR.

So, it is true for parallelograms, but not for proper trapeziums ( ie. where only one pair of parallels)

I haven't investigated the case where PS parallel QR.  You could ry that yourself. 

Bob

Last edited by Bob (2011-10-15 23:29:54)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob