Splitting Masses, Altitudes, Ceva and Menelaus

  1. Splitting mass points using $mP + nP = (m+n)P$ is useful when dealing with transversals. In $\bigtriangleup ABC$, let $E$ be in $\overline{AC}$ such that $AE:EC = 1:2$, let $F$ be in $\overline{BC}$ such that $BF:FC =2:1$, and let $G$ be in $\overline{EF}$ such that $EG:GF=1:2$. Finally, let ray $CG$ intersect $\overline{AB}$ in $D$. Find $CG:GD$ and $AD:DB$.
    Solution: Draw the figure! Assign weight 2 to $C$ and weight 1 to $B$ so that $2C+1B = 3F$. It is now necessary to have weight 6 at $E$ to ``balance" $\overline{EF}$. Since $1C+2A=3E$, we have $2C+4A=6E$, so assign another weight 2 to $C$ for a total weight of 4 at $C$ and assign a weight of 4 to $A$. Then $4A+1B = 5D$. Now the ratios can be read directly from the figure. $CG:GD=5:4$ and $AD:DB=1:4$.
  2. In the previous problem, $AE=EC$, $BF:FC =1:2$, and $EG:GF=2:3$. Show that $CG:GD=17:13$ and $AD:DB=8:9$.
  3. In the previous problem, let $\overline{CD}$ be a median, let $AE:EC=x:1$ and $BF:FC =y:1$. Show that $CG:GD=2:(x+y)$ and $EG:GF=(y+1):(x+1)$.
  4. For an altitude, say $\overline{AD}$ in $\bigtriangleup ABC$, note that $CD\cot B = DB\cot C$. Therefore, assign weights proportional to $\cot B$ and $\cot C$ to $C$ and $B$, respectively. Let $\angle B = 45^\circ$, $\angle C = 60^\circ$, and let the angle bisector of $\angle B$ intersect $\overline{AD}$ in $E$ and $\overline{AC}$ in $F$. Show that $AE:ED = (\frac{\sqrt{3}}{2} +\frac{1}{2}):\sin 75^\circ$ and $BE:EF = (\sin 75^\circ + \frac{\sqrt{3}}{2}): \frac{1}{2}$.
  5. In the previous problem, change $\overline{BF}$ from angle bisector to median. Show that $AE:ED = (3 +\sqrt{3}):3$ and $BE:EF = 2 \sqrt{3}: 1$.
  6. Prove that the altitudes of an acute triangle are concurrent using mass points. Review the clever method of showing this by forming a triangle for which the given triangle is the medial triangle and noticing that the perpendicular bisectors of the large triangle contain the altitudes of the medial triangle.
  7. Let $\bigtriangleup ABC$ be a right triangle with a $30^\circ$ angle at $B$ and a $60^\circ$ angle at $A$. Let $\overline{CD}$ be the altitude to the hypotenuse and let the angle bisector at $B$ intersect $\overline{AC}$ at F and $\overline{CD}$ at $E$. Show that $BE:EF= (3 + 2\sqrt{3}):1$ and $CE:ED= 2:\sqrt{3}$.
  8. Let $\bigtriangleup ABC$ be a right triangle with $AB=17$, $BC=15$, and $CA=8$. Let $\overline{CD}$ be the altitude to the hypotenuse and let the angle bisector at $B$ intersect $\overline{AC}$ at F and $\overline{CD}$ at $E$. Show that $BE:EF= 15:2$ and $CE:ED= 17:15$.
  9. Generalize the previous problems. Let $\bigtriangleup ABC$ be a right triangle with $AB = c$, $BC = a$, and $CA = b$. Let $\overline{CD}$ be the altitude to the hypotenuse and let the angle bisector at $B$ intersect $\overline{AC}$ at F and $\overline{CD}$ at $E$. Show that $BE:EF= a:(a-c)$ and $CE:ED= c:a$.
  10. Prove Ceva's Theorem for cevians that intersect in the interior of the triangle. Three cevians of a triangle are concurrent if and only if the products of the lengths of the non-adjacent sides are equal. (Hint: In $\bigtriangleup ABC$, let $D$, $E$, $F$ be the intersection points of the cevians in sides $\overline{AB}$, $\overline{BC}$ and $\overline{CA}$ , respectively. Let $G$ be the intersection of the cevians, $AD = p$, $DB=q$, $BE= r$, and $EC = s$. Assign weight $sq$ to $A$, $sp$ to $B$, and $rp$ to $C$).
  11. Prove Menelaus' Theorem. If a transversal is drawn across three sides of a triangle (extended if necessary), the product of the non-adjacent lengths are equal.