(1) Triangle $ABC$, inscribed in a circle, has $AB = 15$ and $BC = 25$. A tangent to the circle is drawn at $B$, and a line through $A$ parallel to this tangent intersects $\overline{BC}$ at $D$. Find $DC$.


(2) Let the incircle of triangle $ABC$ be tangent to sides $\overline{BC}$, $\overline{AC}$, and $\overline{AB}$ at $D$, $E$, and $F$, respectively. Prove that triangle $DEF$ is acute.


I have tried several angle and arc formulas... but I can't get anywhere...

  • 1
    $\begingroup$ Hint for (1): look for similar triangles. $\endgroup$ – amd Jul 4 '16 at 18:43
  • $\begingroup$ I am not convinced that lengh DC remains constant whatever be the shape of triangle ABC, but maybe I am wrong... $\endgroup$ – Jean Marie Jul 4 '16 at 20:20
  • $\begingroup$ You mean for all angles at B? $\endgroup$ – Narasimham Jul 4 '16 at 21:24


(1) Triangles $CBA$ and $ABD$ are similar. Exploit that to find $BD$.

(2) The incenter of $ABC$ lies inside $DEF$, because it is the intersection of the angle bisectors.

  • $\begingroup$ I am not doubting your answer but i cant seems to get $\angle BAD = \angle ACB$. How did you get it ?, sorry. $\endgroup$ – A---B Jul 4 '16 at 21:49
  • $\begingroup$ @ritwiksinha $\angle BDA=\angle DBX=\angle CAB$ ($X$ is any point on the tangent away from $A$(downwards) ). Hence your doubt is cleared $\endgroup$ – Qwerty Jul 5 '16 at 4:58
  • $\begingroup$ @ritwiksinha If $E$ is the other intersection between line $AD$ and the circle, then arcs $AB$ and $BE$ are equal. $\endgroup$ – Aretino Jul 5 '16 at 6:41

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