Quadraliteral with strange angles

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Quadraliteral with strange angles

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#1 h Jan 6, 2008, 10:57 PM • 2 Y

Y by Adventure10, Mango247

In a convex quadrilateral $ABCD$ , both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$ . Prove that $AD=CD$ .

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#2 h Jan 6, 2008, 11:18 PM • 2 Y

Y by Adventure10, Mango247

It is clear that A, B and C are points on a circle whose centre is D. Hence DA and DC are of equal length, being two radii of the circle.

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#3 h Jan 6, 2008, 11:55 PM • 2 Y

Y by Adventure10, Mango247

Yes - but does the condition $\angle ADC = 360^{o} - 2(\angle ABC)$ (this is the way you have soled it) is enough to make points $A, B, C$ lay on a circle with center $D$ ? I am not sure, but it seems to be correct...

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#4 h Jan 7, 2008, 12:15 AM • 2 Y

Y by Adventure10, Mango247

In a circle, the angle subtended by a chord at the centre is always twice the angle subtended by the same chord at any point on the circumference on the same side of AB as the centre. The converse is also true: if $\angle$ ABC is twice $\angle$ ABD, where C and D are both on the same side of AB, then a circle can be drawn with centre C passing through D, A and B.

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The QuattoMaster 6000

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The QuattoMaster 6000

#5 h Jan 7, 2008, 2:01 AM • 2 Y

Y by Adventure10, Mango247

Jane_Bennet wrote:

In a circle, the angle subtended by a chord at the centre is always twice the angle subtended by the same chord at any point on the circumference on the same side of AB as the centre. The converse is also true: if $\angle$ ABC is twice $\angle$ ABD, where C and D are both on the same side of AB, then a circle can be drawn with centre C passing through D, A and B.

I don't quite agree with your converse idea. Consider any random points $D$ , $A$ , and $B$ , in the plane. The locus of points $C$ such that $\angle ACB = 2\angle ADB$ is a sector of a circle, but you're claiming that it must be one single point: the circumcenter of $\triangle ABD$ .

Your idea, however, is a good one.
Solution

EDIT: Sorry, Invariant, the sectors of the circles were arc $ADB$ of the circumcircle of $\triangle ABD$ and arc $BDC$ of the circumcircle of $\triangle BCD$ .

This post has been edited 1 time. Last edited by The QuattoMaster 6000, Jan 7, 2008, 11:56 PM

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#6 h Jan 7, 2008, 4:13 AM • 2 Y

Y by Adventure10, Mango247

Here is the theorem that Jane Bennett was thinking about:

If $OA = OB$ and $\angle AOB = 2\angle ACB$ and $C$ , $O$ are in the same half plane wrt. $AB$ , then $O$ is the circumcenter of $\triangle ABC$ .

Note: This problem is easy with complex numbers.

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#7 h Jan 7, 2008, 1:45 PM • 2 Y

Y by Adventure10, Mango247

The QuattoMaster 6000 wrote:

It suffices to show that the locus of $D$ such that $\angle ADB = 2\angle ACB$ and $\angle BDC = 2\angle BAC$ is the circumcenter of $\triangle ABC$ . The locus of $D$ such that $\angle ADB = 2\angle ACB$ is a sector of a circle that passes through $A$ and $B$ and similarly, the locus of $D$ such that $\angle BDC = 2\angle BAC$ is another sector of a circle that passes through $B$ and $C$ . Hence, the locus of $D$ is the intersection(s) between the two sectors of circles described above.

I don't understand. 'The locus of $D$ such that $\angle ADB = 2\angle ACB$ is a sector of a circle that passes through $A$ and $B$ and similarly, the locus of $D$ such that $\angle BDC = 2\angle BAC$ is another sector of a circle that passes through $B$ and $C$ .'
What is this sector of a circle? What is the circumcenter of it? You mean that the locus of $D$ such that $\angle ADB = 2\angle ACB$ is a sector of a circle that passes through $A$ and $B$ - but what is the circumcenter? For sure not $D$ because, to write that above whe should have $DA = DB$ ...

For me the solution should look like this (but I am not sure of it):

$\angle ACB$ and $\angle BAC$ are acute so the circumcenter (let it be $O$ ) of triangle $ABC$ lies inside in $\angle ABC$ . So $\angle ADB = \angle AOB$ thus points $A, B, D, O$ lie on a circle. Similarly I prove that points $B, O, D, C$ lie on a circle (we have 2 different circles). So this circles have three common points $O, D, B$ . $B\neq D$ so it is $D = O$ ; hence $DC = DA$ .
But please, explain me your solutions. Mamy someone manage to do it using isometries?

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WakeUp

#8 h Feb 2, 2011, 4:08 PM • 2 Y

Y by Adventure10, Mango247

Also (later) posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=329925

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#9 h Mar 16, 2018, 4:45 PM • 2 Y

Y by Adventure10, Mango247

Stronger result
In a convex quadrilateral $ABCD$ , both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$ . Prove that points $A,\ B,\ C$ lie on the circle with center at $D$ .

This post has been edited 1 time. Last edited by WolfusA, Mar 16, 2018, 7:27 PM
Reason: Posts 10&12

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#10 h Mar 16, 2018, 5:00 PM • 3 Y

Y by pro_4_ever, Adventure10, Mango247

WolfusA wrote:

Generalized
In a convex quadrilateral $ABCD$ , both $\angle ADB=2\angle ACB$ and $\angle BDC=2\angle BAC$ . Prove that points $A,\ B,\ C$ lie on the circle with center at $D$ .

Isn't that trivial?
That is not a generalization, though, It is just a stronger result

This post has been edited 1 time. Last edited by Mr.Bash, Mar 16, 2018, 5:00 PM

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Arkmmq

#11 h Mar 16, 2018, 6:41 PM • 2 Y

Y by Adventure10, Mango247

Take inversion with center B then we have BC'A'=D'C'A' and D'A'C'=BA'C' so D'C'=BC' and D'A'=BA' then AD=CD...Done

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#12 h Mar 16, 2018, 7:26 PM • 2 Y

Y by Adventure10, Mango247

@Mr.Bash yes, it's rather stronger result.

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