Triangle Heights as Perpendicular Bisectors

[griffin.811]

Hi,

Just wondering, is it necessarily the case that if a height is a perpendicular bisector of the base that the triangle must be an isosceles?

If the above is true, and a height as a perpendicular bisector of a base means that the triangle must then be isosceles, is this only the case if the height is dropped from a right angle vertex, or does this apply for all triangles (right or not), and all angle measurements?

Thanks!

[anshul04]

Yes; if the height is the perpendicular bisector of the base of a triangle, triangle must be an isosceles. And this applies to all triangles (right or not).

You can easily prove it - When you drop the height from an angle to the opposite side of a triangle, it splits the triangle into two triangles (one on either side of this height). Now, in our case, since it's a perpendicular bisector, these two triangles are congruent because
1. height bisects the base of the triangle, and that makes two sides of two triangles equal in length.
2. since height is perpendicular to the base, angle is 90 degrees in both triangles.
3. lastly, the height itself is a common side of both triangles.

So, by SAS (Side-Angle-Side) postulate, two triangles are congruent. And hence, the other two sides of two triangles must also be of same length. That means, our original triangle has two sides of same length and so, is Isosceles.

I hope I made sense in explaining this without drawing a picture.

[griffin.811]

Makes perfect sense. Thanks for the quick response Anshul04!!

[jnelson0612]

Makes perfect sense. Thanks for the quick response Anshul04!!

Yes, thank you!

[RonPurewal]

Hi,

Just wondering, is it necessarily the case that if a height is a perpendicular bisector of the base that the triangle must be an isosceles?

If the above is true, and a height as a perpendicular bisector of a base means that the triangle must then be isosceles, is this only the case if the height is dropped from a right angle vertex, or does this apply for all triangles (right or not), and all angle measurements?

Thanks!

The best way to figure out stuff like this is to sit down and draw a bunch of pictures.

Clearly you know what a "perpendicular bisector" is. So, just draw a bunch of perpendicular bisectors. Then draw in the extra lines to make triangles with those bisectors as heights.

You'll quickly notice that...
"- they'll all be isosceles triangles
"- the vertex angle (from which the height is drawn) can be anywhere from just over 0º (tall skinny triangle) to just under 180º (short stout triangle).