4

$ABC$ is an isosceles triangle, $AB=AC$, and $\measuredangle A=20^{\circ}$. For point $M$ on $AB$ and $N$ on $AC$, we have $AM=NC=BC$. Compute the $\measuredangle BMN$.

In a similar problem (with only one point on one side) the key step was to construct an equilateral triangle on the other side. I tried using that in this problem too, but I didn't manage to solve it.

enter image description here

Michael Rozenberg
  • 194,933
  • What happened to your image? – amWhy Mar 18 '17 at 17:38
  • I fixed the image –  Mar 18 '17 at 17:48
  • Note that $\triangle BCN$ is an isosceles triangle, (because BC = NC) where the angle at $C$ and at $B$ must each be $ \frac 12(180-20) = 80^\circ$. And so $\angle NBC$ and angle $\angle BNC$ must each be $\frac 12(180-80) = 50^\circ$. That means $\angle BNM = 80-50 = 30^\circ.$... – amWhy Mar 18 '17 at 17:49
  • Oops, above I meant that $\angle NBM = 80-50 = 30^\circ$ in the last sentence. – amWhy Mar 18 '17 at 18:09

2 Answers2

4

enter image description here

Locate $D$ such that $\triangle BDC$ forms equilateral triangle.

$\angle ABC = 80^\circ$

$\angle MBD = 20^\circ$

$BD = AM, \angle MBD = \angle NAM, MB = AN$

$\triangle BDM \cong \triangle AMD$

MD = MN

$\triangle MDN$ is an isosceles triangle.

$\angle DMN = 180 - \angle DMB- \angle AMN = \angle MBD = 20$ degrees

$\triangle DCN$ is an isosceles triangle with $20$ degree vertex.

The known angles around $D$ sum to $80^\circ+80^\circ+60^\circ=220^\circ$

The reamianing angle $(\angle MDB) = 140^\circ$

$\angle BMD = 20^\circ$

$\angle BMN = 40^\circ$

user317176
  • 11,017
  • You might want to change your H's to N's, unless you forgot to insert some point H somewhere else. – amWhy Mar 18 '17 at 19:32
  • Was going to ask why MD = MN, but you corrected the argument: their equality doesn't prove the triangles congruent by SSS, but follows from that congruence already proven via SAS. Good. – Edward Porcella Mar 18 '17 at 21:46
0

you can use a $4\times4$ matrix to solve it, where

$$ let ~a = \measuredangle AMN, b = \measuredangle ANM, c = \measuredangle BMN, d = \measuredangle CNM $$

you have

$$ a+c = 180^{\circ} \\ b + d = 180^{\circ} \\ a + b = 160^{\circ} \\ c + d = 200^{\circ} $$

Michael Rozenberg
  • 194,933
HollowMan
  • 53
  • 6