It is known that $\triangle ABC$ is an acute triangle. Let $C'$ be the foott of the perpendicular from $C$ to $AB$, and $D$, $E$ two distinct points on $CC'$. The feet of the perpendiculars from $D$ to $AC$ and $BC$ are $F$ and $G$, respectively. Show that if $DGEF$ is a parallelogram then $ABC$ is isosceles.
From $DF \perp AC$ we get $EG \perp AC$, similarly $EF \perp AB$, so $CE \perp FG$, that is $DE \perp FG$, so DFEG is a diamond.
Then we soon get CC' is also a bisector of $\angle BCA$, that completes the proof.□
k2c901_1 wrote:
Assuming you meant "rhombus" when you said "diamond", I think you're pretty much correct.
Thanks for teaching me an English word of maths.