In a cyclic $\square ABCD$, $BC, CD$ and $DA$ are three tangents of such a circle that its center is on the side $AB$. Proving that $AD + BC = AB$

Question

In a cyclic quadrilateral $ABCD$, $BC, CD$ and $DA$ are three tangents of a circle. The center of the circle is located on the side $AB$. Prove that $$AD + BC = AB$$

Attempt:

First, I thought it to be very easy. So, I let the side $AB$ be the diameter of the large circle by which the quadrilateral $ABCD$ is circumscribed.

So, I made another quadrilateral congruent to $ABCD$ to the opposite side. So, I got a regular hexagon, the one side of which is denoted as $a$ and the radius of large circle = $R$ and the radius of small circle = $r$.

We know that area of regular polygon = $\frac{na^2}{4} \cot \frac{180}{n}$. $ABCD$ is the semi hexagon and so the area of $ABCD =\frac{1}{2}\cdot \frac{6a^2}{4} \cot (\frac{180}{6})^\circ = \frac{3a^2}{4} \cot 30^\circ = \frac{3a^2}{4}\cdot \sqrt3 = \frac {3\sqrt 3a^2}{4}........(i)$

Again $ABCD$ is a trapezium. So, $[ABCD] = \frac{1}{2}(2R + a)\cdot r..........(ii)$

Now from right angled triangle $DJA$: $\frac{r}{a} = \sin 60^\circ$ so $r = \frac{\sqrt 3a}{2}$

So, from equation $(ii)$ again, we get $[ABCD] = \frac{1}{2} (2R + a)\cdot\frac {\sqrt 3a}{2} = \frac{\sqrt 3a}{4} (2R + a)........(iii)$

Now from equation $(i)$ and $(iii)$ we get $\frac{\sqrt 3a}{4} (2R + a) = \frac{3 \sqrt3 a^2}{4}$ and thus $(2R + a) = 3a$ so $2R = 2a$.

Hence, $R = a$. And thus I proved that $2R = a + a \implies AD + BC = AB$.

But that wasn't a satisfactory solution for me in the case of letting $AB$ be the diameter of large circle and I reasonably made it specific. But I am very unaware of the fact that how could I solve that proof for any position of $AB$ such that other three sides of the quadrilateral $ABCD$ are tangents to the small circle?

Thanks in advance.

Source : IMO $1985$

Answer 1

Let $\mathcal{C}$ be a circle wichi is tangent to $BC,CD$ and $DA$ and let it center be $O$. Then $O$ must lie on angle bisector of angle $\angle ADC$ and angle $\angle DCB$. So these angle bisectors meet on $AB$.

Now let $E$ be on $AB$ so that $AE=AD$. We have to prove $BC=BE$ i.e. triangle $BCE$ is isosceles. If $\angle ADE =\alpha$ then $\angle AED =\alpha$ and $\angle ADE =180^{\circ}-2\alpha$. So $\angle BCD =2\alpha$ and thus $\angle OCD =\alpha$. Since $\angle DEB =180^{\circ}-\alpha$ we have $CDEO$ is cyclic.

Let $\angle OEC =\beta$ then $\angle ODC =\beta$ and $\angle ADO = \beta $ so $\angle ABC =180^{\circ}-2\beta$ and thus $\angle BCE =\beta$.