1
$\begingroup$

$\delta$ is the circumscribed circle on a cyclic quadrilateral ABCD. The centre of the inscribed circle of triangle ABC is P, and that of triangle ABD is Q. Let E denote the midpoint of arc BC, and let F denote the midpoint of arc DA of the circle $\delta$.

Prove that the line segment PQ is parallel to EF.

$\endgroup$
3
$\begingroup$

First note the following well known result:

Lemma: Let $ABC$ be a triangle with incenter $I$, and excenter $I_A$ opposite $A$. The line $AII_A$ intersects the circumcircle of triangle $ABC$ at another point $D$ (other than $A$). Then $D$ is the midpoint of arc $BC$ (not containing $A$ of the circumcircle and $DI=DB=DC=DI_A$.

I have provided the proof just in case you do not know this result. Proof: Note that by definition both $I, I_A$ lie on internal angle bisector of $\angle(BAC)$, so $A, I, I_A$ are indeed collinear.

$$\angle ICD =\angle ICB +\angle BCD =\angle ICA +\angle BAD =\angle ICA +\angle CAI =\angle DIC$$

Note that $CI \perp CI_A$, so

$$\angle DCI_A=90^{\circ}-\angle ICD=90^{\circ}-\frac{180^{\circ}-\angle IDC}{2}=\frac{\angle IDC}{2}$$

Thus $DI=DC=DI_A$. Similarly $DI=DB=DI_A$, so we also have $DB=DC$ and $D$ is the midpoint of arc $BC$, so we are done.

Let's continue. Let $G$ be the midpoint of arc $AB$ (not containing $C, D$) By the above lemma, $A, P, E$ are collinear and $PE=PB=PC$, $B, Q, F$ are collinear and $FQ=FA=FC$, $C, P, G$ are collinear and $GP=GA=GB$, $D, Q, G$ are collinear and $GQ=GA=GB$. Note that $GP=GQ$.

For convenience, let us denote the angle subtended by an arc $XY$ as $\angle\widehat{XY}$. We now angle chase.

\begin{align} \angle FQA=\frac{180^{\circ}-\angle \widehat{AGB}}{2} & ; &\angle AQG=\frac{180^{\circ}-\angle \widehat{DFA}}{2} \end{align}

Thus

$$\angle GQB=180^{\circ}-\angle FQA-\angle AQG=\frac{\angle \widehat{AGB}+\angle \widehat{DFA}}{2}=\frac{\angle \widehat{DAB}}{2}$$

$$\angle GQP=\frac{180^{\circ}-\angle \widehat{CD}}{2}=\frac{\angle \widehat{DABC}}{2}$$

$$\angle BQP=\angle GQP-\angle GQB=\frac{\angle \widehat{DABC}}{2}-\frac{\angle \widehat{DAB}}{2}=\frac{\angle \widehat{BEC}}{2}=\angle \widehat{BE}=\angle BFE$$

Thus $PQ \parallel EF$ and we are done.

$\endgroup$
1

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.