A point P is selected anywhere inside an equilateral triangle. What
can you say about the sum of the perpendicular distances from P to
the sides of the triangle? Can you prove your conjecture?
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF.
Similarly the largest equilateral triangle which fits into a circle is LMN and PQR is an equilateral triangle with P and Q on the line LM and R on the circumference of the circle. Show that LM = 3PQ
Prove that the internal angle bisectors of a triangle will never be
perpendicular to each other.
The area of triangle $ABC$ is double the area of $AED$.
$X$ is any suitable point on $AD$
$ZX$ is perpendicular to $AC$, and $ZX$ is equal in length to
So $AXZ$ is an isosceles right-angled triangle.
By sweeping an arc centre $A$ from $X$ to $AZ$ at $N$, $AN$ is
made equal to $AX$
$AN$ to $AZ$ is now in the required ratio.
Drawing from $N$ parallel to $ZC$ the point $D$ is reached.
Because $AND$ and $AZC$ are similar triangles, $AD$ and $AC$ are
in the required ratio.
Excellent and simple!