Proving BD 1/2 AC in a Right-Angled Triangle

This article aims to explore and prove the statement that in a right-angled triangle ABC with B being the right angle and D as the midpoint of AC, the segment BD is equal to one half of AC.

The proof of this statement involves a combination of geometric principles and the application of the Pythagorean theorem. In this article, you will find a step-by-step explanation of the proof.

Given:
- Triangle ABC is right-angled at B.
- D is the midpoint of AC.

To Prove:
- BD 1/2 AC

Proof Using Coordinates and Midpoint Formula

We start by labeling the coordinates of points A, B, and C. For simplicity, let's place point A at (0, 0), point B at (0, b), and point C at (c, 0).

Step 1: Finding the midpoint D
Since D is the midpoint of AC, we can calculate its coordinates as follows:

D ( (0 c) / 2, (0 0) / 2 ) ( c / 2, 0 )

Step 2: Calculating the length AC
Using the distance formula, we can find the length of AC:

AC sqrt( (c - 0)^2 (0 - 0)^2 ) sqrt( c^2 ) c

Step 3: Calculating the length BD
Now, we calculate the distance BD using the distance formula:

BD sqrt( (c/2 - 0)^2 (0 - b)^2 ) sqrt( (c/2)^2 b^2 ) sqrt( c^2/4 b^2 )

Recalling the Pythagorean theorem for right triangle ABC:

AC2 AB2 BC2
implies c2 b2 c2 - b2
implies b2 c2 - b2

Substituting b2 into the expression for BD:

BD sqrt( (c^2/4) (c^2 - c^2)/4 ) sqrt( c^2/4 ) c/2

Thus, we have proven that:

BD 1/2 AC

Using Congruence and the SAS Condition

Alternatively, to establish that BD 1/2 AC, we can use the concept of congruence in similar triangles and the SAS (Side-Angle-Side) criterion.

Suppose BD is not only a bisector but also a perpendicular bisector of AC. In this case, triangles ABD and CBD would be congruent by the SAS criterion, as:

Bisector D makes equal angles at A and C, so angle ABD angle CBD. Segment BD is common to both triangles. Side AD DC since D is the midpoint of AC.

Hence, by the SAS criterion, triangles ABD and CBD are congruent, leading to AB CB and AD DC, thus proving that BD is a bisector of angle B.

Thales' Theorem and Circumcenter

Another way to prove the statement is to use Thales' Theorem and the properties of the circumcenter. If D is the midpoint of AC, and BD bisects angle B, then:

D is the circumcenter of triangle ABC, the point where the perpendicular bisectors of the sides intersect. AC is the diameter of the circumcircle of ABC. By Thales' Theorem, any angle inscribed in a semicircle is a right angle, hence angle ABC 90 degrees.

Since angle ABC is a right angle, and BD bisects angle B, it follows that BD 1/2 AC.

Conclusion

In conclusion, we have proved that in a right-angled triangle ABC with the right angle at B, and if D is the midpoint of AC, then BD is indeed one half of AC. This proof can be approached through the application of the Pythagorean theorem, congruence of triangles, and Thales' Theorem.