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- April 2, 2020 at 5:13 pm #61593Prakash KapadiaParticipant
Let D, E, F be the points of contact of the incircle of an acute-angled triangle ABC with BC, CA, AB respectively. Let I1, I2, I3 be the in-centres of the triangles AF E, BDF, CED, respectively. Prove that the lines I1D, I2E, I3F are concurrent. (RMO 2014, Mumbai Region)

April 2, 2020 at 8:05 pm #61608KOUSHIK SOMParticipantWEe WILL GET BACK TO YOU WITHIN 24 HOURS

April 3, 2020 at 7:17 pm #61873KOUSHIK SOMParticipantObserve that $\angle A F E=\angle A E F=90^{\circ}-A / 2$ and $\angle F D E=\angle A E F=90^{\circ}-A / 2 .$ Again

$\angle E I_{1} F=90^{\circ}+A / 2 .$ Thus

$$

\angle E I_{1} F+\angle F D E=180^{\circ}

$$

Hence $I_{1}$ lies on the incircle. Also

$$

\angle I_{1} F E=(1 / 2) \angle A F E=(1 / 2) \angle A E F=\angle I_{1} E F

$$

Thus $I_{1} E=I_{1} F .$ But then they are equal chords of a circle and so they must subtend equal angles at the circumference. Therefore $\angle I_{1} D F=\angle I_{1} D E$ and so $I_{1} D$ is the internal bisector

of $\angle F D E .$ Similarly we can show that $I_{2} E$ and $I_{3} F$ are internal bisectors of $\angle D E F$ and $\angle D F E$ respectively. Thus the three lines $I_{1} D, I_{2} E, I_{3} F$ are concurrent at the incentre of triangle $D E F$ - AuthorPosts

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