cyclic quadrilateral ...

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Originally Posted by razemsoft21

Can you prove that:

AC is perpendicular to BD

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This is the specific case when angles ADC and ABC are right angles.

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OR

AC divides BD into 2 equal parts

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Again, those triangles would be right angled.

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OR

The angels ACB and ACD are equal ?

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Those angles being equal can have the angles ADC and ABC both $\displaystyle >90^o$

in which case the quadrilateral would not be cyclic,
since opposite angles would not necessarily sum to $\displaystyle 180^o$

Angles ADC and ABC either have to be right angled, or summing to $\displaystyle 180^o$

Having them sum to $\displaystyle 180^o$ is required.

Dear razemsoft,

If you are interested in a formal proof that your quad is a kite i will respond .But first you must go to google to learn what a kite is and its properties.I make no assumptions in this proof but your kite is not cyclic unless you have extra info to prove that angle D =angle B = 90 degrees



bjh

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Originally Posted by bjhopper

Dear razemsoft,

If you are interested in a formal proof that your quad is a kite i will respond .But first you must go to google to learn what a kite is and its properties.I make no assumptions in this proof but your kite is not cyclic unless you have extra info to prove that angle D =angle B = 90 degrees

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Dear bjhopper
I know the properties of the kite very well, and as you know
the question is how to prove that the shape is a kite ?

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Originally Posted by bjhopper

Dear razemsoft,

If you are interested in a formal proof that your quad is a kite i will respond .But first you must go to google to learn what a kite is and its properties.I make no assumptions in this proof but your kite is not cyclic unless you have extra info to prove that angle D =angle B = 90 degrees



bjh

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It has been shown that the quadrilateral can be cyclic even if those angles are not 90 degrees
even if someone refuses to see it.

Proof of kite

Razemsoft's diagram

Connect DB
AC must be the perpendicular bisector of DB in order to produce the two equal x angles at A

AD=AB A is on the perp bisector
AC=AC x=x AD=AB
Triangles ADC and ABC are congruent SAS
Angle D = Angle B CPCT
Kite is defined


bjh

Quote:

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Originally Posted by bjhopper

Proof of kite

Razemsoft's diagram

Connect DB
AC must be the perpendicular bisector of DB in order to produce the two equal x angles at A

The above line is false

AD=AB A is on the perp bisector
AC=AC x=x AD=AB
Triangles ADC and ABC are congruent SAS
Angle D = Angle B CPCT
Kite is defined


bjh

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Please reference emakarov's diagram earlier in the thread.

I get the point finnally.

bjh