In equilateral triangle ABC with AB = 4 cm, points M on AC and N on BC satisfy CM = BN, and neither point is a vertex. Lines AN and BM meet at G. (a) Draw CH perpendicular to AB at H and find CH. (b) Prove AN = BM and find the angle AGM.

Step-by-step solution

1. 1

   (a) Use the fact that in an equilateral triangle the altitude is also a median. Because $CA=CB$, the foot $H$ of the altitude from $C$ is the midpoint of $AB$, so

   $AH=HB=\frac{AB}{2}=2\ \text{cm}$

   This is what lets you turn one triangle into a right triangle with two known sides.

2. 2

   (a) Apply Pythagoras in right triangle $ACH$.

   $CH^{2}=AC^{2}-AH^{2}=4^{2}-2^{2}=16-4=12$

   $CH=\sqrt{12}=2\sqrt3\approx3.46\ \text{cm}$

   This is the familiar $\frac{\sqrt3}{2}\times\text{side}$ result for an equilateral triangle, here $\frac{\sqrt3}{2}\times 4$.

3. 3

   (b) Find the right pair of triangles to compare. The tempting pair $\triangle ABN$ and $\triangle CAM$ does not work, because $AM=4-CM$ is generally different from $BN$. Compare $\triangle ABN$ and $\triangle BCM$ instead:

   $AB=BC=4\ \text{(equilateral)},\qquad \angle ABN=\angle BCM=60^\circ,\qquad BN=CM\ \text{(given)}$

   The included angle sits between the two matched sides in each triangle, so this is a genuine SAS match.

4. 4

   (b) Conclude the equality of the two cevians. By SAS, $\triangle ABN\cong\triangle BCM$, and corresponding parts give

   $AN=BM$

   as well as the angle equality we need next: $\angle BAN=\angle CBM$.

5. 5

   (b) Chase the angles at the crossing point $G$. Since $M$ lies on $AC$, the ray $BM$ splits the $60^\circ$ angle at $B$:

   $\angle ABM=60^\circ-\angle CBM$

   Adding the equal angle from the congruence,

   $\angle BAN+\angle ABM=\angle CBM+\left(60^\circ-\angle CBM\right)=60^\circ$

   In triangle $ABG$ the angles at $A$ and $B$ are exactly these two, so

   $\angle AGB=180^\circ-60^\circ=120^\circ$

6. 6

   (b) Convert to the angle asked for. $G$ lies on segment $BM$, so $\angle AGM$ and $\angle AGB$ are supplementary:

   $\angle AGM=180^\circ-120^\circ=60^\circ$

   Notice the answer does not depend on where $M$ and $N$ sit: sliding them along the sides (keeping $CM=BN$) rotates the pair of cevians together and always leaves $60^\circ$ between them. A coordinate check with $A(0,0)$, $B(4,0)$, $C(2,2\sqrt3)$ and $CM=BN=2$ gives $\angle AGM=60^\circ$, confirming it.