In a triangle $ABC$, the angle bisector of $A$ and the cevians $BD$ and $CE$ concur at a point $P$ inside the triangle. Show that the quadrilateral $ADPE$ has an incircle if and only if $AB=AC$.
Problem
Source: Central American Olympiad 2007, Problem 2
Tags: geometry, symmetry, angle bisector
dyaka
12.06.2007 09:26
The center of an inscribed circle is the intersection of angle bisectors. Consider a circle and an arbitrary point outside of the circle. Draw tangents from the point to the circle. Then select an arbitrary point on the circle and draw the tangent. So far the whole thing is symmetric. The tangent crosses the bisector (the first line). From the intersection, draw the other tangent. Due to symmetry the resulting quadrilateral has pairs of adjacent equal sides. More symmetry, and triangle ABC is isosceles.
tchebytchev
19.12.2007 16:25
suppose that $ (AP)$ meet $ BC$ at $ Q$. $ \bullet :$ if $ ADPE$ has an incircle . we have $ \widehat{APE}=\widehat{APD}$ so $ \widehat{APB}=\widehat{APE}+\widehat{EPB}=\widehat{APD}+\widehat{DPC}=\widehat{APC}$ and $ \widehat{PAB}=\widehat{PAC}=\frac{\widehat{A}}{2}$ so $ \widehat{PBE}=\widehat{PCD}$ so $ \triangle PEB \sim \triangle PDC$ $ \Longrightarrow \frac{CD}{BE}=\frac{PD}{PE} (\star) .$ in other hand $ \frac{BP}{PD}=\frac{AB}{AD}$ and $ \frac{PE}{PC}=\frac{AE}{AC}$ but since $ \frac{BP}{PC}=\frac{BQ}{QC}=\frac{AB}{AC}$, we have $ \frac{PE}{PD}=\frac{AE}{AD} (\star \star) .$ and by ceva theorem's $ 1=\frac{BQ}{QC}\frac{DC}{AD}\frac{AE}{EB}=\frac{AB}{AC}\frac{DC}{AD}\frac{AE}{EB}$ and with $ (\star )$ and $ (\star \star)$ we have $ AB=AC$. $ \bullet:$if$ AB=AC$ then $ 1=\frac{BQ}{QC}\frac{DC}{AD}\frac{AE}{EB}=\frac{AB}{AC}\frac{DC}{AD}\frac{AE}{EB}$ $ \Longrightarrow \frac{DC}{AD}=\frac{BE}{EA}$ $ \Longrightarrow \frac{AC-AD}{AD}=\frac{AB-EB}{EA}$ $ \Longrightarrow AD=AE$ and $ AP=AP$ and $ \widehat{PAE}=\widehat{PAD}$ so $ PE =PD$ and in this case $ AD+EP=AE+PD$ so $ ADPE$ has an incircle .