I'll talk you through it so you can see why it's true, and then
you can set up the 2-column proof on your own:

Look at the two pointy triangles, hanging down like moth-wings
on each side of 'OC'.

-- Their long sides are equal, OA = OB, because both of those lines
are radii of the big circle.

-- Their short sides are equal, OC = OC, because they're both the same line.

-- The angle between their long side and short side ... the two angles up at 'O',
are equal, because OC is the bisector of the whole angle there.

-- So now you have what I think you call 'SAS' ... two sides and the included angle of one triangle equal to two sides and the included angle of another triangle.
(When I was in high school geometry, this was not called 'SAS' ... the alphabet
did not extend as far as 'S' yet, and we had to call this congruence theorem
"broken arrow".)

These triangles are not congruent the way they are now, because one is
the mirror image of the other one. But if you folded the paper along 'OC',
or if you cut one triangle out and turn it over, it would exactly lie on top of
the other one, and they would be congruent.

So their angles at 'A' and at 'B' are also equal ... those are the angles that
you need to prove equal.

5 0

4 years ago

Plz plz plz help me with this

AnnZ [28]

Answer:

m∠A = 65°

Step-by-step explanation:

Opposite angles of the quadrilateral add up to 180.

This means x + 148 = 180 -> x = 32.

m∠A = 2x + 1 -> 2(32) + 1

m∠A = 65°.

4 0

3 years ago