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**sydbernard****Member**- Registered: 2017-09-04
- Posts: 6

Give an example that disproves the proposition “If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary.” Is the converse proposition true? I've only been able to come up with examples that prove this proposition and can't think of any that show it's false.

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,354

hi sydbernard,

Welcome to the forum.

A is the common vertex, ABC is one of the angles; ADE the second; and the bisectors meet at F.

It didn't take long to construct this diagram; having some geometry software makes it easy. I use Geometer's Sketchpad which will cost you money. There's a similar program called Geogebra which is free. You'll find a download here:

Hope that helps,

Bob

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

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**sydbernard****Member**- Registered: 2017-09-04
- Posts: 6

Thank you!

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**thickhead****Member**- Registered: 2016-04-16
- Posts: 1,086

When we say vertex of an angle, as far as my knowledge goes,it is the point where the two arms meet. The two additional points taken on the arms are only for nomenclature. This way I do not approve bob bundy's approach. I feel the statement “If the bisectors of two angles with a common vertex are perpendicular, then the angles are supplementary.”needs additional clause. "with a common side." We may have angle ABC and angle DAF with their bisectors perpendicular but the two angles are not supplementary. I am sorry for not showing it on a diagram.

**{1}Vasudhaiva Kutumakam.{The whole Universe is a family.}(2)Yatra naaryasthu poojyanthe Ramanthe tatra Devataha{Gods rejoice at those places where ladies are respected.}**

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,354

Hhhmm. A diagram would be useful here. I've given this some more thought and the main lesson we can claim from this question is the importance for the setter in making very clear what the question is. I've come up with three alternatives in addition to the one above:

In the first the vertex is A and the angles ABC and ACB are the base angles of a triangle and are bisected.

Let's suppose ABC = 2x and ACB = 2y, and that the bisectors cross at 90.

Then, in triangle BDE we have angles of x, 90 and y and so in triangle CDE angle BEC = x.

So angle BEA = 180 - x

and in triangle ABE angle A = 0.

So in this case the bisectors cannot cross at 90.

In my second diagram the vertex is at G and angles FGH and FGJ are bisected.

If FGH = 2x and FGJ = 2y, then 2x + 2y + 2x + 2y = 360, => x+y = 90, and the angles are supplementary.

For my third diagram the angles meet at a point and share a common line. If the angles are 2x and 2y and we know that x+y = 90, then 2x + 2y = 180 and the angles are supplementary.

??????

Maybe I should adapt the third so that the angles do not share a common line. Then the angles would not be supplementary. I leave this fourth diagram to the interested reader.

Bob

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

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