Robin S.
asked • 04/30/17Prove it isosceles
If the angular bisector of an triangle bisects the opposite side, prove that the other two sides are equal.
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3 Answers By Expert Tutors
Inactive Tutor answered • 04/30/17
Tutor
New to Wyzant
//Update to my answer below//
I uploaded a proof of the theorem and its application to this problem if you would care to look at it. Its in my resources.
//original answer below//
In this case the ratio of the two segments is 1, so the ratio of the lengths of the other two sides of the triangle is also 1, i.e. they are equal and the triangle is isosceles. A geometric proof requires that you extend the bisector and then draw a line from the end of the bisector back to one of the other vertices of the triangle that is parallel to the side of the triangle opposite the vertex to which you draw the line. I can't do that here but you can look up the theorem online for a method. Doing this will enable you to prove that the bisector intersects the side at right-angles.
The result follows from the angle-bisector theorem: the ratio of the lengths of the two segments of a triangle's side formed by a line that bisects the opposite angle equals the ratio of the lengths of the other two sides of the triangle.
In this case the ratio of the two segments is 1, so the ratio of the lengths of the other two sides of the triangle is also 1, i.e. they are equal and the triangle is isosceles. A geometric proof requires that you extend the bisector and then draw a line from the end of the bisector back to one of the other vertices of the triangle that is parallel to the side of the triangle opposite the vertex to which you draw the line. I can't do that here but you can look up the theorem online for a method. Doing this will enable you to prove that the bisector intersects the side at right-angles.
Inactive Tutor answered • 04/30/17
Tutor
New to Wyzant
I believe that Patrick is correct.
You have the bisected angle (making two equal angles), and the bisector side, and the bisected opposite side; make the drawing; you will see that these two triangles have only 'SSA' characteristics and so NO PROOF OF CONGRUENCE.
//this has been revised from my earlier version//
Inactive Tutor
I respectfully and professionally disagree with Kenneth.
You MAY NOT ASSUME that the angles formed by the bisector and the opposite side form equal angles.
Given triangle ABC, draw the angle bisector from B to AC which bisects AC and angle B.
Label the intersection of the bisector to AC point M
Segment MB is congruent to itself by reflexive.
Segment AM and MC are congruent per the bisector.
Angle ABM and Angle CBM are congruent.
THAT's IT!!!
YOU MAY NOT ASSUME ANYTHING about angles BMA and BMC
You need either another pair of congruent sides, another pair of congruent angles,
or the additional stipulation that the bisector is really the PERPENDICULAR BISECTOR
which would make BMA and BMC both right angles.
I have the drawing saved. We are confusing the student.
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04/30/17
Inactive Tutor
But what if the base of the original triangle is essentially tilted, say, down to the right, thereby making the right side of the triangle longer than the left. (It's possible that I'm missing the strict definition of opposite here...) Does that illustrate Patrick's point, or am I barking up the wrong tree?
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04/30/17
Arthur D.
tutor
If a line segment bisects an angle and bisects its opposite side, then the line segment must be perpendicular to the line that it bisects. If the line segment bisects the opposite side, and is not perpendicular to it, then the line segment can't bisect the angle opposite the bisected side.
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04/30/17
Arthur D.
tutor
A line segment can't bisect an angle and bisect its opposite side at the same time and not be perpendicular to the bisected side.
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04/30/17
Inactive Tutor
Thank you Kenneth and THANK YOU DAMIAN for supplying the theorem that this poor student desperately needs.
Ya'll are the BOMB! WE ALL learned something today!
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04/30/17
Inactive Tutor
ARTHUR, if you draw isosceles triangle ABM with BA = BC, and draw angle bisector BM (M being midpoint of AC), then indeed BM is perpendicular to AC at M.
But now go back to your drawing, and de-isoscelize the triangle by tilting AC by rotating it slightly (with M as the center of rotation). NOW the angle bisector is no longer perpendicular to the rotated AC, but the angle B still has been bisected and side AC is still bisected. But the triangle isn't isosceles, obviously.
I think that Patrick AGREES.
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04/30/17
Arthur D.
tutor
Hi Kenneth, I drew isosceles triangle ABC and drew the angle bisector BM where M is the midpoint of AC. I rotated AC clockwise using a compass. I opened the compass as wide as the distance AM. I put the point of the compass at M and rotated the compass until the arc hit the line segment AB. Now segment AC is a slanted line going downward from left to right. I no longer have a triangle unless I rotate, which was segment BC, clockwise and extend it so that BCA is now again a triangle. Or I can say that when I rotated AC, I pulled and extended BC so that I would still have a triangle.
BC got pulled clockwise to the left. Now the angle ABC is not bisected. Angle ABM is not equal in measure to angle MBC.
Point M is still the midpoint of segment AC but angle ABC is no longer bisected. When you tilt AC, you must tilt BC which makes the angle ABC no longer bisected. Please comment on my reasoning. Thanks, Kenneth.
Arthur D.
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04/30/17
Inactive Tutor answered • 04/30/17
Tutor
New to Wyzant
NO
They are not.
You need another pair of congruent angles or another pair of congruent sides
OR IF YOU SAID the angular bisector is actually the PERPENDICULAR bisector of the
opposite side, then it bisects the opposite side AND they meet at right angles, which
are the same.
But you didn't
so it's not
Robin S.
I am not that much qualified as much you people are, but as far as my little mind can think, in my opinion Sir Patrick is right
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05/01/17
Inactive Tutor
Hi Robin
I uploaded a proof of the theorem and its application to this problem if you would care to look at it. Its in my resources.
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05/01/17
Robin S.
Thanx Damien sir. This helps. But if you could write it as an answer and not a comment I would click 'this answers my question'.
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05/01/17
Inactive Tutor
OK. Sorry, I'm new at this.
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05/01/17
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Inactive Tutor
04/30/17