Let \(ABC\) be a right-angled triangle with \(\angle B=90^{\circ}\). Let \(I\) be the incentre if \(ABC\). Extend \(AI\) and \(CI\); let them intersect \(BC\) in \(D\) and \(AB\) in \(E\) respectively. Draw a line perpendicular to \(AI\) at \(I\) to meet \(AC\) in \(J\), draw a line perpendicular to \(CI\) at \(I\) to meet \(AC\) at \(K\). Suppose \(DJ=EK\). Prove that \(BA=BC\).
Problem
Source: RMO 2016 Karnataka Region P5
Tags: geometry
khyeon
16.10.2016 14:47
where's $J$?
svatejas
16.10.2016 14:52
Sorry typo.
khyeon
16.10.2016 14:57
I think $DJ=DK$ impossible $IJ=IK$ possible....
svatejas
16.10.2016 14:58
Sorry another typo now its fixed.
Ayushakj
16.10.2016 15:00
Add complete paper, indeed. photo of paper.
Ayushakj
16.10.2016 15:27
angle JIK=45.
geekpradd
16.10.2016 15:46
Here's how I did. Please check this. We note that \(\angle AIC=135^{\circ}\). This gives \(\angle AIK = \angle AIE = 45^{\circ}\). Hence $AI$ is also angle bisector of \(\angle KIE\). Similarly $CI$ bisects \(\angle JID\). This yields that $AKI$ is congruent to $AEI$ and $CJI$ is congruent to $CDI$ by $ASA$. This shows $KEI$ and $JDI$ are isosceles. Since $KE$ = $JD$, $KEI$ and $JDI$ are congruent. Let $KE$ intersect $AI$ at $X$ and $JD$ intersect $CI$ at $Y$. It then follows that \(\angle AXK\) and \(\angle CYJ\) are right angles. Then we can show that $\angle AKX$ and \(\angle CJY\) are equal after some manipulation. Hence the triangles $AKX$ and $CJY$ are similar \(\angle XAK\) $=$ \(\angle YCJ\) which gives that \(\angle A\) $ = $ \(\angle B\)
Tuhin_wemath3
19.10.2016 19:53
The same qstn in rmo wb 2016
math_and_me
03.10.2018 05:02
http://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2016/11/3sol.pdf