finekiss

Well i found a very cute solution of the following problem in geometry:

Prove that if two bisectors in any triangle are equal the triangle is isosceles.

Solution:

Let ABC be any triangle with BD=CF where (BD) and (CF) are the bisectors of angle <ABC and <ACB respectively. Show that ABC is isosceles. (Assume that <ABC = 2b, <ACB = 2c and <BAC = 2a )

• Construct triangle EBD such that <EBA = c and <BDE = b+c

• Construct triangle GFC such that <GCA = b and <GFC = b+c

We then have:

<BED + <EBD + <EDB = 180 = 2a + 2b +2c

<BED +(b+c)+(b+c) = 2a +2b + 2c

That imply <BED = 2a. Similarly, <FGC = 2a

The quadrilaterals EADB and AGCF are inscribed since <BED = <BAD and <FAC = FGC

So <EAB + <BAD + (b+c) = 2a + 2b + 2c

<EAB = b+c, Similarly <CAG = b+c

Then <CAG = <EAB, Thus < EAB + <BAC + <CAG = (b+c) + 2a + (b+c) = 2a+2b+2c = 180

Therefore E, A, and G are Collinear.

In triangle ACG we have:

<C + <A + <G = 2a + 2b +2c

b + b + c + <G = 2a +2b + 2c That imply <G = c

In the quadrilatreral EGCB :

<EGC + <EBC = (2a+b) +(2b+c) = 2a + 2b + 2c = 180

So EGCB = is inscribed, but EB = CG since triangles EBD and GFC are congruent.

Therefore EGCB is an isosceles trapezoid and so <BAE = <ABC that imply b+c = 2b which means c =b[/url]

Prove that if two bisectors in any triangle are equal the triangle is isosceles.

Solution:

Let ABC be any triangle with BD=CF where (BD) and (CF) are the bisectors of angle <ABC and <ACB respectively. Show that ABC is isosceles. (Assume that <ABC = 2b, <ACB = 2c and <BAC = 2a )

• Construct triangle EBD such that <EBA = c and <BDE = b+c

• Construct triangle GFC such that <GCA = b and <GFC = b+c

We then have:

<BED + <EBD + <EDB = 180 = 2a + 2b +2c

<BED +(b+c)+(b+c) = 2a +2b + 2c

That imply <BED = 2a. Similarly, <FGC = 2a

The quadrilaterals EADB and AGCF are inscribed since <BED = <BAD and <FAC = FGC

So <EAB + <BAD + (b+c) = 2a + 2b + 2c

<EAB = b+c, Similarly <CAG = b+c

Then <CAG = <EAB, Thus < EAB + <BAC + <CAG = (b+c) + 2a + (b+c) = 2a+2b+2c = 180

Therefore E, A, and G are Collinear.

In triangle ACG we have:

<C + <A + <G = 2a + 2b +2c

b + b + c + <G = 2a +2b + 2c That imply <G = c

In the quadrilatreral EGCB :

<EGC + <EBC = (2a+b) +(2b+c) = 2a + 2b + 2c = 180

So EGCB = is inscribed, but EB = CG since triangles EBD and GFC are congruent.

Therefore EGCB is an isosceles trapezoid and so <BAE = <ABC that imply b+c = 2b which means c =b[/url]