Triangle construction!?

Question

How we can construct a triangle if we know the length of two sides - say b and c - and the length of bisector between them (the bisector of angle A) ?

Answer 1

The previous answer was on the right track but is wrong in two places. First, the last two equations are wrong. Second, the solution procedure, once you have got three equations, was not quite straightforward, if not incorrect.

I will use the similar convention that the previous answer has used. Her first equation was right, so I repeat here (using a, b, d to simplify the notation though):

BM = CM * b / c

Assume s is half of angle A, d is the length of the bisector, the remaining two equations are:

c^2 + d^2 - 2cd * cos(s) = BM^2
b^2 + d^2 -2bd * cos(s) = CM^2

To solve them, you want to remove cos(s) from the equations. You multiply the first by b, second by d, subtract second equation from the first, you get (after quite some simplifications):

c^2(d^2-bc) = (b^2 + c^2 + ab) CM^2

You get CM here. Then BM is easy to find as BM = CM * b / c. Then you should be able to get cos(s), then s easily.

Once you know BM and CM, the next question is how to draw it.
Draw a horizontal line whose length is BM + CM. From point B, draw a circle with length c; from point C, draw a circle with length b. The two circle should intercept at point A.

Answer 2

The bisector will form a right angle between the unknown side and both of the other sides. Then you can use the pythagorean theorem to find the the length of the unknown side:

so: a^2 + b^2 = c^2 (or in your case c^2 - b^2 = a^2) where "c" is one of your known line segments and "b" is the length of the bisector line.

solve for a and then DOUBLE it because the "a" you are actually solving for is just the length of the line segment between the bisector line and one of your known lines. Since the bisector line "cuts" the angle in half, it will also cut your unknown line in half.

Now you have the length of each of the lines in your triangle.

Hope this helps.

Answer 3

When you say 'the length of the bisector' do you mean the length for point A to the opposite side?

I don't think, as some have suggested, that the bisector is necessarily perpendicular to the opposite side unless, of course, sides b and c are equal in length.

At the moment, I don't see how to construct the triangle.

Answer 4

There might be a more elegant, no calculation method, but this is what I came up with:

Say the triangle is ABC and the 2 angles formed A1 and A2.
Say the bisector is AM.

Then AM/sin B = MB/ sin A1 = AB/ sin M1
AM/sin C = MC/sin A2 = AC/ sin M2

From here: MB/AB = MC/AC

Also BM^2 = AB^2 +AM^2 - 2* AB * AM * sinA1
MC^2 = AC^2 + AM^2 - 2*AC* AM * sin A1
So now you have 3 equations with 3 unknowns BM, MC and A1
Square MB/AB = MC/AC and you can easily find sin A1.
I got: sin A1 = AM.(AC + AB)/(2*AB*AC)
So now you have sin A/2 , there are some formulas to find sin A.
Now you build a triangle knowing 2 sides and the sin of the angle.
I suggest you draw for instance side b, then you make a circle with side b as its diameter. The ABC triangle has a height equal to b * sin A, build it. All is easy from here.

PS: I supposed A is <90. The other case(s) should be very similar.

RESPONSE TO STANYAN: I PUT SIN INSTEAD OF COS I'M SORRY.
BUT MY EQUATION SOLVING IS FINE, CHECK YOURS.

LOOK:

MB^2 AB^2
------- = ----------
MC^2 AC^2

So: AB^2 AB^2 + AM^2 - 2AB*AM*cosA1
--------- = ----------------------------------------------
AC^2 AC^2 + AM^2 - 2AC*AM*cosA1

AB^2*AC^2+AB^2*AM^2-2AC*AM*AB^2*cosA1= AB^2*AC^2+AM^2*AC^2-2AB*AM*AC^2*cosA1

By simplifying we get:

AB^2*AM- 2AC*AB^2*cosA1 = AC^2*AM- 2AB*AC*cosA1

So (AB^2-AC^2)*AM
---------------------- = cos A1,
2AB*AC*(AB-AC)


So cos A1 = AM*(AB+AC)/2AB*AC, if AB<>AC.

EVERYTHING ELSE I'VE SAID FROM HERE ON CAN EASILY BE ADAPTED TO A COS INSTEAD OF SIN.

I've seen DIMOS answer geometry questions I don't think all these explanations were necessary.
Besides he probably would have liked a classical greek construction, based on the fact that the bisector is equally distanced from the sides. I'm sorry I can't think of any.

BY THE WAY YOU CAN USE THE COSINE OF THE HALF ANGLE FOR A SIMILAR CONSTRUCTION.

Answer 5

draw a line the length of bisector
construct a perpendicular to it
using compass, find intersection of the 2 sides
connect the 3 points