Is the inverse proposition that the bisector of the two base angles of an isosceles triangle is equ

Solution: Swap the conditions and conclusions of a proposition to get its inverse proposition Therefore, the inverse proposition of the proposition "the bisector of the two base angles of an isosceles triangle is equal" is "In a triangle, if the angle bisector of the two angles is equal, then the triangle is an isosceles triangle".

It is a true proposition

This is the famous Steiner-Lemmers theorem.

Two types of evidence. In abc, be, cf are bisectors of b, c, be=cf. Verification: ab=ac

If you set ab≠ac, you might as well set ab>ac, so that acb > abc, so that bcf = fce= acb 2> abc 2= cbe= ebf.

In BCF and CBE, because BC=BC, BE=CF, BCF> CBE

So bf>ce. (1)

To make a parallelogram begf, then ebf= fgc, eg=bf, fg=be=cf, and cg, so fcg is an isosceles triangle, so fcg= fgc.

Because FCE > FGE, ECG < EGC.

Therefore, ce>eg=bf (2)

Obviously, (1) and (2) are contradictory, so assuming that ab≠ac is not true, then there must be ab=ac.

Argument 2 In ABC, assuming B C, then a little F can be taken on CF', so that f'be=∠ecf', which has cf cf'。

Extend BF'Handed over to AC to A', then by ba'e=∠ca'f', there is δa'be∽δa'cf'.

Thus a'b/a'c=be/cf'≥be/cf=1.

Then in a'B.C., by A'b≥a'c, got:

a'cb≥∠a'bc, i.e., c ( b + c) 2, hence b c.

Then let's assume b c, i.e., there is b = c.

So abc is an isosceles triangle.

Because the question of the original proposition is: "A triangle is an isosceles triangle", and the conclusion of the trace is "the two base angles of this triangle are equal", so the inverse proposition of the proposition "the two base angles of an isosceles triangle are equal" is "A triangle with two equal angles is an isosceles triangle".

<> known: In ABC, ab=ac, bf, ce, respectively, abc, the angular bisector of acb, acb

Verification: bf=ce, that is, the bisector of the two base angles of the isosceles triangle is equal.

Proof: ab=ac, empty old.

abc= acb, bf, and ce, respectively, are the angular bisector of abc, acb, bce= cbf, abc= acb, bc=bc, bce cbf, bf=ce, that is, the ascending lines of the flat bucket family at the two bottom angles of the isosceles triangle spike are equal

Suppose that two rays of two base angles are emitted at 30 degrees of their respective angles, and conclusion: if two rays are shot in a straight line at the same velocity, but must be in a vacuum, then the bisector of the two base angles is equal, but there is no specific value.

Prove that two triangles consisting of waist lines and angular bisector lines are congruent.

Because the top angle is a common angle, the waist line is the same as the two triangles, one side is also equal, the bisect angle is also equal, and the corners are equal.

Two triangles congruence.

So the angular bisector is also equal.

The original proposition is set up: "a triangle is an isosceles triangle", the conclusion is "this triangle has equal two base angles", and the inverse proposition of the proposition "the two base angles of an isosceles triangle are equal" is "two triangles with equal base angles are isosceles triangles".

So the answer is: a triangle with two equal angles is an isosceles triangle

A triangle with two equal angles is an isosceles triangle [Analysis] Analysis: First find the question and conclusion of the original proposition, and then replace the question and conclusion with each other, and then get the inverse proposition of the original proposition. Because the original proposition is set up as:

A triangle is an isosceles triangle", and the conclusion is: "The two base angles of this triangle are equal", so the proposition "The two base angles of an isosceles triangle are equal." The inverse proposition of the group attack model is:

A triangle with two equal angles is an isosceles triangle. Difficulty] Average.

A triangle with two bottom bisectors equal is an isosceles triangle using congruence proof, first proving that the outer pair is congruent, then the corresponding angles are equal, and the angles with the large bisector are also equal, and finally using the equiangular equisides