ABCD is cyclic with BC=CD

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Sayan

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Sayan

#1 h May 24, 2014, 10:33 AM • 2 Y

Y by Adventure10, Mango247

Suppose $ABCD$ is a cyclic quadrilateral with $BC = CD$ . Let $\omega$ be the circle with center $C$ tangential to the side $BD$ . Let $I$ be the centre of the incircle of triangle $ABD$ . Prove that the straight line passing through $I$ , which is parallel to $AB$ , touches the circle $\omega$ .

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utkarshgupta

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utkarshgupta

#2 h May 24, 2014, 2:22 PM • 1 Y

Y by Adventure10

Do you mean tangent by touches the circle $\omega$ cause I think it cannot be a tangent(only in a specific case).

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utkarshgupta

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utkarshgupta

#3 h May 24, 2014, 5:36 PM • 1 Y

Y by Adventure10

I have some proof(that it intersects) which I hope is correct.
I am giving a partial solution right now and will complete when I have time.

Obviously $AIC$ is a straight line.
Also it is pretty well known $CD=CI=CB$ .
Let the circle passing through $D,I,B$ with $C$ centre be $\omega_1$

Now consider the homothety $H_C:I \to A$ that is centred at $C$ which takes $I$ to $A$ .
We will show $H_C:\omega \to \omega_1$ .

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PROF65

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PROF65

#4 h May 24, 2014, 8:50 PM • 2 Y

Y by Adventure10, Mango247

consider $I_e$ the excenter ,T the projection of C on the parallel to AB through I
X= $\frac{R-r}{2},tan \frac{A}{2}=\frac{r}{s-a}=\frac{R}{s}=\frac{r'}{\frac{a}{2}}$ ,where[a=BD,s= $\frac{BD+AB+AD}{2}$ ] the consequence is obvious

Attachments:

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inthedarkness

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inthedarkness

#5 h May 25, 2014, 12:50 PM • 2 Y

Y by Adventure10, Mango247

Firstly, $A,I,C$ are colinear and $BC=IC=DC$ And let $R$ be the tangent point of $\omega$ with $BD$ . Clearly, $BR=RD$ and now let $Q$ be the pedal of projection from $C$ to the straight line through $I$ and parallel to $AB$ .
Now, consider, $\triangle IQC$ and $\triangle BRC$ : We have: $BC=IC$ , $\angle CBR=\angle QIC=\angle A/2$ . And now as they are right angled so, $\angle BCR=\angle ICQ$ .so they are congruent and finally, $CQ=CR$

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nvk_khanglun

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nvk_khanglun

#6 h May 29, 2014, 7:31 AM • 2 Y

Y by Adventure10, Mango247

I solved this problem,but i don't know Latex

inthedarkness wrote:

Firstly, $A,I,C$ are colinear and $BC=IC=DC$ And let $R$ be the tangent point of $\omega$ with $BD$ . Clearly, $BR=RD$ and now let $Q$ be the pedal of projection from $C$ to the straight line through $I$ and parallel to $AB$ .
Now, consider, $\triangle IQC$ and $\triangle BRC$ : We have: $BC=IC$ , $\angle CBR=\angle QIC=\angle A/2$ . And now as they are right angled so, $\angle BCR=\angle ICQ$ .so they are congruent and finally, $CQ=CR$

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inthedarkness

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inthedarkness

#7 h May 29, 2014, 8:19 AM • 1 Y

Y by Adventure10

So your method is same as me? Congrats!

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inthedarkness

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inthedarkness

#8 h May 29, 2014, 8:22 AM • 1 Y

Y by Adventure10

So your method is same as me? Congrats!

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MrRTI

191 posts

MrRTI

#9 h May 29, 2014, 12:15 PM • 2 Y

Y by primesarespecial, Adventure10

Let $IX$ be the tangent from $I$ to the circle $\omega$ and $X$ lies on $\omega$ and let $\omega$ touches $BD$ at $M$ .
The equality $CB = CI = CD$ is well-known and easily proven, then the triangles $\triangle{CXI} \cong ruent \triangle{CMB}$ . So, $\angle{CAB} = \angle{CBM} = \angle{CIX}$ . So, $IX$ is parallel to $AB$ . Thus, $IX$ is the line through $I$ and parallel to $AB$ .

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