Points cyclic iff angle is right

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Points cyclic iff angle is right

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Source: RMO Maharashtra and Goa 2016, P4

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#1 h Oct 11, 2016, 7:42 AM • 3 Y

Y by Adventure10, Mango247, Rounak_iitr

Let $\triangle ABC$ be scalene, with $BC$ as the largest side. Let $D$ be the foot of the perpendicular from $A$ on side $BC$ . Let points $K,L$ be chosen on the lines $AB$ and $AC$ respectively, such that $D$ is the midpoint of segment $KL$ . Prove that the points $B,K,C,L$ are concyclic if and only if $\angle BAC=90^{\circ}$ .

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#2 h Oct 11, 2016, 11:00 AM • 2 Y

Y by Adventure10, Mango247

First Let $B,K,C,L$ be con cyclic..
Extend $DC$ to point $P$ such that $DC=DP.$
And then we get $PAC$ is isosceles and $PKCL$ is parallelogram .Also we get $KPAD$ is a cyclic quadrilateral.
And then it is very easy

Proving it other way is quite obvious..

This post has been edited 3 times. Last edited by phymaths, Oct 11, 2016, 11:01 AM
Reason: Typo

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#3 h Oct 11, 2016, 11:11 AM • 2 Y

Y by Adventure10, Mango247

This is an interesting problem.

Let $B,K,C,L$ lie on a circle. Let $AD$ meet $(ABC)$ again at $X$ and note that $DA\cdot DX=DB\cdot DC=DK\cdot DL$ and the points $A,K,X,L$ are concyclic. As $X$ is the Miquel's Point of $BCKL$ , $\triangle XKL \sim \triangle XBC$ and it follows that $X,D,B,K$ and $X,D,C,L$ are concyclic. Thus, $AX$ is a diameter in $(AKL)$ and a median in triangle $AKL$ ; and as $AKL$ is non-isosceles, it follows that $D$ is the circumcenter of $AKL$ and so $\angle BAC=90^{\circ}$ . All steps are reversible and the converse promptly holds.

This post has been edited 1 time. Last edited by anantmudgal09, Oct 11, 2016, 11:12 AM

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#4 h Oct 11, 2016, 11:23 AM • 2 Y

Y by kapilpavase, Adventure10

If we go by considering $BKCL$ concyclic , just observe that $KL$ is the $A$ symmedian of $ABC$ . Also as $AD$ is perpendicular to $BC$ , by the theorem :- "In triangle $ABC$ , if $AD$ is the altitude from $A$ to $BC$ , and $O$ is circumcenter of $ABC$ , then, angle $BAD$ = angle $CAO$ "... Using this property , we get that the circumcenter of $ABC$ lies on the $A$ median of the triangle $ABC$ . But this happens iff $ABC$ is either an equilateral triangle , an isosceles triangle , or a right angled triangle. As $ABC$ is given to be a scalene triangle , this implies angle $BAC = 90$ . Proving the other way round , i.e taking angle $A = 90$ is trivial by angle chasing .

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#5 h Oct 11, 2016, 1:05 PM • 3 Y

Y by gauss01, Adventure10, Mango247

anantmudgal09 wrote:

This is an interesting problem.

Yeah, interesting indeed .
Suppose, $BKCL$ is concyclic
Let $\angle ALK$ = $k$ . Then, $\angle KBC$ = $(180-k)$ . for this, $\angle ABD$ = $180$ - $\angle KBC$ = $k$ ., $\angle DAK$ = $\angle DAB$ = $90$ - $\angle ABD$ = $90-k$ ......(@)
Let, $O$ is the circumcentre of $\triangle AKL$ . Here, $\triangle OAK$ is isosceles. Here, $\angle AOK$ =2 ( $\angle AKL$ )= $2k$ & $\angle OAK$ = $90 -k$ ......(b)
From, @ & (b) , $\angle DAK$ = $\angle OAK$ , SO, $O$ lies on line $AD$ . If we construct a perpendicular bisector of $KL$ through $D$ , it will go through $O$ . So, $O$ & $D$ is same point. so, $KL$ is a diameter of $\odot AKL$ . AS, $\angle KAL$ is inscribed in a semi-circle, so, $\angle KAL$ = $\angle BAL$ = $\angle BAC$ = $90^{\circ}$ .

This post has been edited 6 times. Last edited by raknum007, Feb 1, 2017, 1:24 PM

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#6 h Dec 10, 2016, 9:05 PM • 3 Y

Y by Abhinandan18, Adventure10, Mango247

Sorry for reviving, but for the only if part I have got a trigonometric solution without any construction....So posting it.
Suppose, $BKCL$ be cyclic. Let $\angle ACB= a$ . $\implies$ $\angle DAC= (\pi\ /2 - a)$ . Let $\angle KLC= b$ $\implies$ $\angle KBC= b$ $\implies$ $\angle ABC= (\pi\ - b)$ $\implies$ $\angle BAD= (b- \pi\ /2)$ . Now, apllying Sine Rule in $\triangle ADK$ and $\triangle ADL$ , we get $AD / KD = -sin a/cos b$ and $AD / DL = sin b/ cos a$ respectively. $\implies\ sin 2a + sin 2b=0 \implies\ 2sin(a+b)sin(a-b)=0$ . Now, as $\triangle ABC$ is scalene, $\angle ABC \neq\ \angle ACB$ . So, $a+b\neq\ \pi$ . So, $cos (a-b)=0$ . So, $|a-b|=\pi\ /2$ . Now, as $BC$ is largest side, $\angle ABC$ and $\ angle ACB$ both must be acute, hence, $\angle KBC=b \ > \pi\ /2 > \angle ACB=a$ . So, $b-a=\pi\ /2$ . so $a= b - \pi\ /2$ , hence, $\angle DKA= \angle DAK$ . So, $D$ is the circumcenter of $\triangle AKL$ .Hence $\angle BAC=\pi\ /2$ .

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#7 h Dec 10, 2016, 11:09 PM • 3 Y

Y by dgmath97, Adventure10, Mango247

If part:

In $\Delta ABC$ if $\angle BAC=\frac{\pi}{2}$ , then $AD^2=BD.DC$ , and $AD$ being the median in right angled $\Delta AKL$ we have $AD^2=KD^2=KD.DL$ (since $AD=KD=DL$ ), and we get $BD.DC=KD.DL$ i.e. $\frac{BD}{KD}=\frac{DL}{DC}$ , which means that the two triangles $\Delta BKD$ and $\Delta LCD$ are similar. This gives $\angle BKD=\angle LCD$ .

Only If part:

If $BKCL$ is concyclic, then $\angle AKL=\angle ACB$ and $\angle LAB$ being the common angle, $\Delta AKL$ and $\Delta ABC$ are similar.

Drawing the median $AM$ in $\Delta ABC$ , so that $M$ is the midpoint of $BC$ , we have that $AM$ and $AD$ play the same corresponding roles (of the median from the equal angle $A$ ) in the two similar triangles, and $AC$ and $AK$ play the same corresponding roles (of being the sides opposite to equal angles $\angle ALK=\angle ABC$ , since the two are similar.)

So, the angle between $AM$ and $AC$ in $\Delta ABC$ must be equal to the angle between $AD$ and $AK$ in $\Delta AKL$ .

In particular, $\angle MAC=\angle DAB$ , which means $AM$ (median from $A$ in $\Delta ABC$ ) is the isogonal conjugate of $AD$ (altitude from $A$ in $\Delta ABC$ ) in $\Delta ABC$ .

But we also know, that the isogonal conjugate of $AD$ is $AO$ where $O$ is the circumcenter of $\Delta ABC$ . So, $AO$ and $AM$ is the same line, which means that essentially $M$ and $O$ are the same point. So $M$ is the circumcenter of $\Delta ABC$ , which means that $\angle BAC= \frac{\pi}{2}$

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