It looks easy

Geometry Level 2

Given isosceles triangle $ABC$ , with $\angle ABC= \angle ACB=50°$ and $AB=AC$ . $M$ is a point inside triangle $ABC$ whether $\angle MBC=10°$ and $\angle MCB=20°$ . What is the measure of $\angle AMB$ (in degrees)?

P/s: Try to use pure geometry!

Inspiration .

The image isn't accurate!

The answer is 80.

1 solution

Huân Lê Quang
Oct 1, 2015

Draw the bisector of MCB, expand it to let it cut MB at F

We can easily prove that triangle AFB and triangle AFC are congruent and they are isosceles triangles with FAB = FAC = 40 degrees

Call M' which is the intersection of the bisector of FAC and CM => M'AC= 20 degrees and M'CA = 30 degrees.

Call X which is the intersection of the bisector of FAM' and CM' => triangle AXC is an isosceles triangle and AXC = 120 degrees.

Can easily prove that AXF = XFC = 120 degrees

Triangle AXF and triangle AXM' are congruent (FAX = M'AX = 10 degrees; AF is a common side; AXF = AXC = 120 degrees)

=> AF = AM'

=> triangle AFM' is isosceles with AFM' = AM' F = (180-20)/2 = 80 degrees

=>AM' F + AFB = BFM' = 80 + 100 = 180 degrees (because FBA = FAB = 40 degrees => AFB = 100 degrees)

=>B,F,M' are collinear

=> BF cut CM at M'

but BF also cut CM at M and BF isn't coincident with CM

=> M is coincident with M'

=> angle AMB is coincident with angle AM' F

=> AMB = 80 degrees

It looks easy

The answer is 80.

1 solution

0 pending reports