Geometry Problem?

Question

Suppose you have a triangle ABC as follows: pick a point D on AB and draw a line segment from D to C: DC; pick a point E on AC and draw a line segment from E to B: EB; label the point where these two segments intersect F.

Let the following be true:

1. angle EBC = angle DCB
2. Line segment BD = line segment CE

Show that ABC is isosceles.

Answer 1

The answer above doesn't provide a proof that BEC and BCD are the same... which is the whole trick of the proof.

So, by point 1, the triangle BFC is isosceles (since two of its angles are the same, the two sides BF and CF are of equal length).
By the law of sines, we know that EC and BD must be equal as well (since the opposite angles are equal).

So let the angle ABC = a + x, and let the angle ACB = a + y, where a is that angle that they have in common from above. Let angle BDC = b, let angle BEC = c. Then b + x = y + c and since we showed before that the line BD is the same length as line EC, we know angle b and c are equal, which tells us that x and y are equal too. Therefore, angle ABC = a+x = a+y = angle ACB, and the triangle is isosceles.

Answer 2

Triangles BEC and CDB are the same.
Therefore, triangle ABC is isoceles, since angles ABC and ACB are the same. QED

Addendum: Well, as you see with ya_tusik's answer, it took a PhD in math to prove this.

Triangles BEC and CDB are the same because
1) angle EBC = DCB
2) Line segment BD = line segment CE
3) Line segment BC = line segment BC !

Addendum: Dan Lobos has given a correct proof, so why the thumbs down?
Meanwhile, ya_tusik's proof does not make any mention of the necessary fact that BD = CE, so how did she prove that angles b and c are the same?

I vote for Dan's proof as the only exactly correct proof here so far. Mine has left out a special case in the matter of "equal triangles", which would force a slightly longer proof. Lobo's proof is the shortest.

Answer 3

1. Since EBC = DCB, FBC is isocles, which means that FB = FC
2. Since BD = CE and statement 1. are true, DF = CF
3. Since 1. and 2. are true, triangles DFB and CEF are same (perhaps not the right term, but their sides and angles are equal)
4. Since 3. is true, angles ABC and ACB are the same, which means that the triangle ABC is isoceles

As always, a sketch helps a lot when solving problems like this.

Answer 4

I bear in mind that there is a thorem that if the two aspects of a traingle are equivalent then the attitude opposite to them are additionally comparable. Please see the thorem. So the bisectors make a isoscele traingle with the facet in between the angles have been getting bisected.i'll later write you the information of the theorm. because of the fact the each and each a million/2 of the angles are equivalent, then the entire angles are equivalent. right here you got here upon the two angles are equivalent. As consistent with definition: in a traingle if the two angles are equivalent, then that traingle is stated as a isoscele traingle.