Problem 1309

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The midpoint of the diameter $AB$ is the center $O$ of the circle containing $C$ and $D$, so $OC = OD$: $O$ is on the perpendicular bisector of $[CD]$. If we call $M$ the midpoint of $[CD]$, then $(OM)$ is precisely the perpendicular bisector of $[CD]$, and it remains to prove that the midpoint $G$ of $[EF]$ is also on this perpendicular bisector of $[CD]$, hence that $GC = GD$. Since $C$ and $D$ are on the circle with diameter $[AB]$, $\widehat{ACB} = \widehat{ADB} = 90^\circ$. It follows that $\widehat{FCE}$ and $\widehat{FDE}$ are also right angles, so $C$ and $D$ also belong to the circle with diameter $[EF]$ and center $G$, hence $GC = GD$, which completes the proof.

## - Area of a Triangle -

The area of a triangle can be calculated in several ways. The main one is: area = $\frac{1}{2}$ Base $\times$ Height, considering that a triangle has three bases and three heights. This is the basis for the proof of the Pythagorean theorem in Euclid's Elements. The two shaded triangles are congruent: it suffices to rotate $\triangle AEB$ by $90^\circ$ around $A$ to obtain $\triangle ACF$. They therefore have the same area. For $\triangle AEB$, we choose $AE$ as the base, and the corresponding height is equal to $CA$ since $CB$ is parallel to the base $EA$, so the area is half the area of the square $ACDE$. For $\triangle ACF$, we choose $AF$ as the base, and the corresponding height is equal to $HA$ if $CH$ is perpendicular to $AB$, hence parallel to the base $AF$, and the area is equal to half the area of the rectangle $AF GH$. Similarly (but Euclid repeats the proof a second time), half the area of the square $BCJK$ is equal to half the area of the rectangle $BIGH$. Therefore, the sum of the areas of the two squares is equal to the sum of the areas of the two rectangles, which is the area of the square of the hypotenuse.

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If we denote $a, b, c$ as the three lengths $BC, CA, AB$, and $\widehat{A}, \widehat{B}, \widehat{C}$ as the three angles $\widehat{BAC}, \widehat{CBA}, \widehat{ACB}$, the area $S$ can also be written as: $S = \frac{1}{2} ab \sin \widehat{C} = \frac{1}{2} bc \sin \widehat{A} = \frac{1}{2} ca \sin \widehat{B}$. It suffices to note, for example, that the height opposite $BC = a$ has a length of: $AC \cdot \sin \widehat{C} = b \sin \widehat{C} = AB \cdot \sin \widehat{B} = c \sin \widehat{B}$.

Another formula allows calculating the area of a triangle using only the lengths $a, b, c$ of the sides $BC, CA, AB$. This is Heron's formula:

$$S = \frac{1}{4} \sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$$

First, note that the area of the triangle is zero if and only if the three vertices are collinear, so one of the sides is the sum of the other two, which means that $(a+b-c)(a-b+c)(-a+b+c)$ is zero. However, the proof requires grouping the terms in pairs: $(a+b+c)(a+b-c) = (a+b)^2 - c^2 = (a^2 + b^2 - c^2) + 2ab$, and $(a-b+c)(-a+b+c) = c^2 - (a-b)^2 = (c^2 - a^2 - b^2) + 2ab$. It remains to use the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos \widehat{C}$: $(a+b+c)(a+b-c) = 2ab(1 + \cos \widehat{C})$ and $(a-b+c)(-a+b+c) = 2ab(1 - \cos \widehat{C})$ have a product of: $4a^2b^2(1 - \cos^2 \widehat{C}) = 4a^2b^2 \sin^2 \widehat{C}$, whose square root: $2ab \sin \widehat{C}$ is indeed four times the area. Another way to write Heron's formula is to call $p$ the semiperimeter: $p = \frac{a+b+c}{2}$. We then have: $S = \sqrt{p(p-a)(p-b)(p-c)}$.

## - Angles -

We recall that the sum of the angles in a triangle is $180^\circ$. A useful way to write this is: if we extend one side, $AB$ for example, beyond $A$, the exterior angle, supplementary to $\widehat{A}$, is the sum of the two base angles: $\widehat{B} + \widehat{C}$.

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We also recall that an isosceles triangle is a triangle with two equal sides and two equal angles: if it has two equal sides, it has two equal angles, and vice versa. Finally, the angle bisector divides an angle into two equal angles. Points on the angle bisector are equidistant from the sides of the angle. It follows that the three angle bisectors of a triangle intersect at a point $I$: if we call $I$ the intersection of the two angle bisectors of $\widehat{B}$ and $\widehat{C}$, $I$ is at the same distance from $AB$ and $BC$, from $BC$ and $AC$, and therefore from $AC$ and $AB$, which implies that it is on the angle bisector of $\widehat{A}$. If we call $r$ the distance from $I$ to the three sides, the circle with center $I$ and radius $r$ is tangent to the three sides: it is called the inscribed circle of the triangle, and $I$ is often referred to as the center of the inscribed circle of the triangle.