# Triangle geometry problem

Problem: Given a right triangle $ABC$ with $A=90°$. On $AC$ we label $D$ such that $\angle ABD=(1/3)\angle ABC$. On AB we label $E$ such that $\angle ACE=(1/3)\angle ACB$. $F$ is the intersection of $BD$ and $CE$. The angle bisector of BFC and FBC meet at $I$. Prove that $DIE$ is an isosceles triangle.

So far I've find out that $\angle BFC=120°$, so $\angle BFI=\angle IFC=60°$. I cannot figure out any pairs of equal triangle.

Note: I have learnt equal triangle. But I haven't learnt trigonometry, of similar triangle or any property of circle or quadrilateral.

• It's not true. Checked in GeoGebra. May be $\angle ABD=(1/3) \angle ABC$? – bigant146 Feb 15 '17 at 16:25
• Maybe $\angle ACE = \frac{1}{3}\angle ABC$? (otherwise I don't understand how you could get that $\angle BFC = 120^{\circ}$). – kolobokish Feb 15 '17 at 16:28
• If so you can investigate quadrilateral $FICD$ and find that $FI=FD$ – bigant146 Feb 15 '17 at 16:32
• Yes, \angle ABD$= 1/3 \angle ABC$. Sorry for the typo – Lê Đức Minh Feb 15 '17 at 16:35
• @bigant146 I've solve the problem! Thank you very much! – Lê Đức Minh Feb 15 '17 at 18:39

Start formatting your posts!

1) $I$ is the intersection point of all three interior angle bisectors of triangle $BCF$;

2) $\angle\, DFI = 60^{\circ} = \angle \, CFD$;

3) $\angle \, FCI = \angle \, FCD$;

4) Triangles $DFC$ and $IFC$ are congruent by 2) and 3);

5) By 4) $DF = IF$;

6) Analogous arguments yield that triangles $EBF$ and $IBF$ are congruent;

7) By 6) $EF = IF$;

8) By 5) and 7) $DF = IF = EF$ and moreover $\angle \, DFI = 120^{\circ} = \angle \, EFI$

9) By 8) triangles $DEI$ is equilateral and thus $DI = EI$.