# 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 $AD + BC = AB$

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?

Source : IMO $$1985$$

• You assumed that $CD$ is parallel to $AB$. Why? Commented Mar 7, 2019 at 18:36
• @Vasya It is very clear that the small circle is inscribed in the hexagon and so it is regular. And that's why $CD \parallel AB$. Another fact is that $AB$ is the diameter of the large circle. Commented Mar 7, 2019 at 18:39
• This is of course not true Commented Mar 7, 2019 at 18:41
• Is this again regional competition in Bangladesh? Commented Mar 8, 2019 at 14:39

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 $$\angle BCE =\beta$$.