ISL 2005

[shehab ahmed]

In a $\triangle ABC$ satisfying $AB + BC = 3AC$ the incircle has centre I and touches the sides $AB$ and $BC$ at $D$ and $E$ respectively.Let $K$ and $L$ be the symmetric points of $D$ and $E$ with $I$.Prove that the quadrilateral $ACKL$ is cyclic.

[zadid xcalibured]

MY SOLUTION:

$a+c=3b$ so $s=\frac{a+b+c}{2} \Rightarrow s=2b.$let the extension of $AL$ meet $BC$ at $M$.then$CM=s-b=b=AC$.so we can determine $\angle MAC=90-\frac{C} {2}=\angle AMC=90-\angle MLE.\angle KLE=\frac{B} {2}.\angle KLM=\angle MLE+\angle KLE=\angle ACD$

[Tahmid Hasan]

MY SOLUTION:

$a+c=3b$ so $s=\frac{a+b+c}{2} \rightarrow s=2b.$let the extension of $AL$ meet $BC$ at $M$.then$CM=s-b=b=AC$.so we can determine $\angle MAC=90^{\circ}-\frac{C}{2}=\angle AMC=90^{\circ}-\angle MLE.\angle KLE=\frac{B}{2}.\angle KLM=\angle MLE+\angle KLE=\angle ACD .$

nice use of homothety ,but there are problems with latex.

[zadid xcalibured]

ঐ, Homothety কই পাইলি?

[Phlembac Adib Hasan]

My solution is similar to Jdd vaia.I'll talk about it with him on Friday's PMS class.Actually the second lemma of Zhao's sheet kills this one.(And also, this problem was the first of the related problems of that lemma.)