Regulated quadrilateral

Stun · Poincare Conjecture Offline Joined: 11 Oct 2004 Posts: 116

Given ABCD is a regulated quadrilateral inscribed in a circle (O) (which has $AB\cdot CD=BC\cdot DA$ ).
Let K, H be the midpoints of AC and BD respectedly. Prove that AK+CK=BH+DH.

[Moderator edit: LaTeXed the equation. Please use the standard notation $x\cdot y$ for the product of two numbers x and y, rather than the obsolete notation x.y (note that LaTeX has the command \cdot for a dot like $\cdot$ ).]

**Posted:** Sun Jan 16, 2005 5:53 am

RaMlaF · Hodge Conjecture Offline Joined: 17 Jul 2003 Posts: 83

That is equivalent to prove that $AC=BD$ , which is not true (at least it doesn't seem to be )... or perhaps you mean that we have to prove $AH+CH=BK+DK$ .

**Posted:** Sun Jan 16, 2005 6:10 am

grobber

He's referring to a harmonic cyclic quadrilateral. It has lots of nice properties. I'm sure he meant $AH+CH=BK+DK$ , as RaMlaF said.

I think similar problems have been discussed before. For instance, I'll use the fact that $AC,BD$ bisect the angles $\angle BKD,\angle AHC$ without proof (take it as a nice problem Smile

). From this we can deduce that it suffices to show that the chords obtained by producing $AH,DK$ are equal, which, in turn, would be derived from $\angle DAH=\angle ADK\ (*)$ , so let's focus on proving $(*)$ .

Now I'll use another property of the harmonic quadrilaterals: the tangents through $A,C$ to the circle concur in $P\in BD$ , and the symmetric situation also holds. From here we find $K$ to be the image of $O$ , the center of the circle, through the inversion of pole $P$ and power $PB^2=PD^2$ , so $BDKO$ is cyclic, meaning that (look at the second equality) $\angle DKA=\frac{\angle DKB}2=\frac{\angle DOB}2=\angle DCB$ , and since $\angle DAK=\angle DBC$ , we find $DKA,DCB$ to be similar. In the same way we show that $AHD,ABC$ are similar, so $\angle DAH=\angle CAB=\angle BDC=\angle ADK$ , which is precisely $(*)$ .

Stun · Poincare Conjecture Offline Joined: 11 Oct 2004 Posts: 116

Sorry for my stupid mistake. I really mean AH+CH=BK+DK Mr. Green

**Posted:** Sun Jan 16, 2005 3:27 pm

Stun · Poincare Conjecture Offline Joined: 11 Oct 2004 Posts: 116

But Grober, can you explain to me what we should do after we have $\angle DAH=\angle ADK$ ? Confused

**Posted:** Sun Jan 16, 2005 5:25 pm

RaMlaF · Hodge Conjecture Offline Joined: 17 Jul 2003 Posts: 83

I don't see so clearly what to do with $\angle DAH = \angle ADK$ ... but i have another solution (without magic properties ... Very Happy

just kiddin'):
Let $D'$ be the reflection of $D$ with respect to $A$ . Then $\angle BAD = \angle BCD$ and $\frac {BA}{AD'} = \frac{BA}{AD}=\frac{BC}{CD}$ since $ABCD$ is an harmonic quadrilateral. Therefor $BAD'\simeq BCD$ , with $BD'=\frac{BD.AB}{BC}$ And, since $DH=BH$ and $DA=DA'$ we have $BD'=2AH$ so $AH=\frac{BD.AB}{2BC}$ . Similarly, $CH=\frac{BD.BC}{2AB}$ . And doing the same with $BK$ and $DK$ the problem reduces to show that $BD(\frac{AB}{BC}+\frac{BC}{AB})=AC(\frac{BC}{CD}+\frac{CD}{BC}) \iff BD(\frac{AB}{BC}+\frac{CD}{AD})=AC(\frac{AB}{AD}+\frac{C...$ (because $ABCD$ is harmonic) $\iff BD(AB.AD+BC.CD)=AC(AB.BC+CD.AD) \iff \frac{AC}{BD}=\frac{AB.AD+BC.CD}{AB.BC+CD.AD}$ . And we see that $(AB.AD+BC.CD).\sin BCD = (AB.BC+CD.AD).\sin CDA = 2.Area(ABCD)$ . Using that, and the Extended Law of Sines with $\sin BCD=\frac{BD}{2R}$ and so on, we obtain the desired result.

**Posted:** Sun Jan 16, 2005 6:33 pm

grobber

Since $\angle DAH=\angle DAK$ , let $P,Q$ be the intersections of $DK,AH$ with the circle. Then $DPQA$ is an isosceles trapezoid, meaning that $DK,AH$ produced form equal chords of the circle. This is what we wanted.

**Posted:** Sun Jan 16, 2005 8:10 pm

Stun · Poincare Conjecture Offline Joined: 11 Oct 2004 Posts: 116

YEAH, I finally find another simply solutions Mr. Green

Use the condition $AB\cdot CD=AD\cdot BC$ and the Ptoleme theorem we have the triangle DAH and BAC are similar.
$DH=\frac{2R\cdot S(ABD)}{AC\cdot BD}$
And qith the same way we will get the result
Sorry if I did make mistakes because I must be very hurry now Mr. Green

**Posted:** Sun Jan 16, 2005 9:03 pm

blue allrise · New Member Offline Joined: 13 Jan 2005 Posts: 5

this problem is not difficult
to prove the equality you can use this
the tangent to the circle at B,D and the line AC are concurrent after you use ptoleme theorem; Mr. Green

**Posted:** Wed Jan 26, 2005 3:59 am