Question #7208

How do you prove that "The altitude to the base of an isosceles triangle is also a median"?

Expert's answer

Let ABC be an isosceles triangle with base AC, so AB=BC.

Let also BD be the

altitude.

We should prove that BD is also a median, so AD=DC.

Consider

triangles ABD and CBD.

Then we have the following:

1) Since ABC is

isosceles, we have that angle A = angle C.

2) The angles BDA and BDC are

right and so they are equal

3) It follows from 1) and 2) that the third

angles ABD and CBD are equal as well.

3) AB=CB

5) the side BD is

common for both triangles.

Thus due to 3), 4) and 5) thriangles ABD and

CBD are equal, since they have two equal sides and equal angles

between

them.

In particular, AD=CD, and so BD is median.

Let also BD be the

altitude.

We should prove that BD is also a median, so AD=DC.

Consider

triangles ABD and CBD.

Then we have the following:

1) Since ABC is

isosceles, we have that angle A = angle C.

2) The angles BDA and BDC are

right and so they are equal

3) It follows from 1) and 2) that the third

angles ABD and CBD are equal as well.

3) AB=CB

5) the side BD is

common for both triangles.

Thus due to 3), 4) and 5) thriangles ABD and

CBD are equal, since they have two equal sides and equal angles

between

them.

In particular, AD=CD, and so BD is median.

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