Subfield: Geometry


Question: Let $A B C D$ be a convex quadrilateral. A circle passing through the points $A$ and $D$ and a circle passing through the points $B$ and $C$ are externally tangent at a point $P$ inside the quadrilateral. Suppose that

$$
\angle P A B+\angle P D C \leq 90^{\circ} \quad \text { and } \quad \angle P B A+\angle P C D \leq 90^{\circ} \text {. }
$$

Prove that $A B+C D \geq B C+A D$.

Solution:
We start with a preliminary observation. Let $T$ be a point inside the quadrilateral $A B C D$. Then:



$$
\text{Circles $(B C T)$ and $(D A T)$ are tangent at $T$}
\newline
\text { if and only if } \angle A D T+\angle B C T=\angle A T B \text {. } \tag{1}
$$

Indeed, if the two circles touch each other then their common tangent at $T$ intersects the segment $A B$ at a point $Z$, and so $\angle A D T=\angle A T Z, \angle B C T=\angle B T Z$, by the tangent-chord theorem. Thus $\angle A D T+\angle B C T=\angle A T Z+\angle B T Z=\angle A T B$.

And conversely, if $\angle A D T+\angle B C T=\angle A T B$ then one can draw from $T$ a ray $T Z$ with $Z$ on $A B$ so that $\angle A D T=\angle A T Z, \angle B C T=\angle B T Z$. The first of these equalities implies that $T Z$ is tangent to the circle $(D A T)$; by the second equality, $T Z$ is tangent to the circle $(B C T)$, so the two circles are tangent at $T$.

<img_3708>

So the equivalence (1) is settled. It will be used later on. Now pass to the actual solution. Its key idea is to introduce the circumcircles of triangles $A B P$ and $C D P$ and to consider their second intersection $Q$ (assume for the moment that they indeed meet at two distinct points $P$ and $Q$ ).

Since the point $A$ lies outside the circle $(B C P)$, we have $\angle B C P+\angle B A P<180^{\circ}$. Therefore the point $C$ lies outside the circle $(A B P)$. Analogously, $D$ also lies outside that circle. It follows that $P$ and $Q$ lie on the same $\operatorname{arc} C D$ of the circle $(B C P)$.

<img_3746>



By symmetry, $P$ and $Q$ lie on the same arc $A B$ of the circle $(A B P)$. Thus the point $Q$ lies either inside the angle $B P C$ or inside the angle $A P D$. Without loss of generality assume that $Q$ lies inside the angle $B P C$. Then

$$
\angle A Q D=\angle P Q A+\angle P Q D=\angle P B A+\angle P C D \leq 90^{\circ}\tag{2}
$$

by the condition of the problem.

In the cyclic quadrilaterals $A P Q B$ and $D P Q C$, the angles at vertices $A$ and $D$ are acute. So their angles at $Q$ are obtuse. This implies that $Q$ lies not only inside the angle $B P C$ but in fact inside the triangle $B P C$, hence also inside the quadrilateral $A B C D$.

Now an argument similar to that used in deriving (2) shows that

$$
\angle B Q C=\angle P A B+\angle P D C \leq 90^{\circ} .\tag{3}
$$

Moreover, since $\angle P C Q=\angle P D Q$, we get

$$
\angle A D Q+\angle B C Q=\angle A D P+\angle P D Q+\angle B C P-\angle P C Q=\angle A D P+\angle B C P .
$$

The last sum is equal to $\angle A P B$, according to the observation (1) applied to $T=P$. And because $\angle A P B=\angle A Q B$, we obtain

$$
\angle A D Q+\angle B C Q=\angle A Q B
$$

Applying now (1) to $T=Q$ we conclude that the circles $(B C Q)$ and $(D A Q)$ are externally tangent at $Q$. (We have assumed $P \neq Q$; but if $P=Q$ then the last conclusion holds trivially.)

Finally consider the halfdiscs with diameters $B C$ and $D A$ constructed inwardly to the quadrilateral $A B C D$. They have centres at $M$ and $N$, the midpoints of $B C$ and $D A$ respectively. In view of (2) and (3), these two halfdiscs lie entirely inside the circles $(B Q C)$ and $(A Q D)$; and since these circles are tangent, the two halfdiscs cannot overlap. Hence $M N \geq \frac{1}{2} B C+\frac{1}{2} D A$.

On the other hand, since $\overrightarrow{M N}=\frac{1}{2}(\overrightarrow{B A}+\overrightarrow{C D})$, we have $M N \leq \frac{1}{2}(A B+C D)$. Thus indeed $A B+C D \geq B C+D A$, as claimed.