Christmas Streak 72/88: Geometry Rush #2

Geometry Level 3

Given a triangle $\rm ABC$ as shown on the right, a point $\rm D$ is chosen such that $\overline{\rm BD}//\overline{\rm AC}$ and $\angle \rm ADB=90^{\circ}.$

$\overline{\rm AB}=2$ and $\overline{\rm CE}=4.$

$\angle \rm BAC=114^{\circ}.$ Find the size of $\angle \rm ABC$ in degrees.

This problem is a part of <Christmas Streak 2017> series .

42 45 43 44 40 41

1 solution

Boi (보이)
Dec 16, 2017

(All angle sizes are in degrees.)

As shown on the right, the midpoint of $\overline{\rm BC},$ namely $\rm M,$ is the key to this problem.

$\overline{\rm AM}=2,$ since $\rm M$ is the circumcenter of $\triangle \rm ACE.$

Let $\angle {\rm ABC}= x,$ then we get $\angle {\rm AMC} = x.$

Then $\angle {\rm MAC} = \dfrac{x}{2}.$ And we know that $\angle {\rm BAM} =180- 2x.$

Since $\angle {\rm BAC} = \dfrac{x}{2} + 180 - 2x = 114,$

$\dfrac{3}{2}x=66~\Leftrightarrow~x = \boxed{44}.$

$\angle {\rm AMC}=180-x$

A Former Brilliant Member - 3 years, 5 months ago

Christmas Streak 72/88: Geometry Rush #2

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