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.

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.