In a triangle ABC , the internal bisectors of angles B and C meet at P and the external bisectors of the angles B and C meet at Q. prove that:
angle BPC + angle BQC = 2 rt. angles.
To prove that angle BPC + angle BQC is equal to two right angles, we first need to establish some properties of the angles in question. Let's start by labeling the given diagram:
Given:
Triangle ABC with internal bisectors of angles B and C meeting at point P,
and external bisectors of angles B and C meeting at point Q.
Now, let's recall some properties of angle bisectors in a triangle:
1. The internal bisector of an angle divides the opposite side of the triangle into two segments that are proportional to the adjacent sides.
2. The external bisector of an angle extends the opposite side of the triangle.
Using these properties, we can establish the following relationships:
1. Since P is the internal bisector of angle B, it divides side AC into two segments, AP and PC, such that (AB / BC) = (AP / PC).
2. Similarly, since P is also the internal bisector of angle C, it divides side AB into two segments, AP and PB, such that (AC / BC) = (AP / PB).
3. From properties 1 and 2 above, we can conclude that (AB / BC) = (AP / PC) = (AC / BC) = (AP / PB).
Therefore, AP / PC = AC / BC = AP / PB, which implies that AC = PC and AC = PB.
Now, let's consider the angles in question:
4. Angle BPC is the angle formed by sides PB and PC.
5. Angle BQC is the angle formed by sides BQ and QC.
6. Notice that angle BPC and angle BQC share the side BC.
Using the fact that triangle ABC is a closed figure, we can say that:
7. The sum of the measures of all angles in a triangle is always equal to 180 degrees.
Therefore, angle ABC + angle BAC + angle CAB = 180 degrees.
Since angle BPC and angle BQC are adjacent angles that are on the same line (BC), their measures must add up to 180 degrees:
8. Angle BPC + angle BQC = 180 degrees
We have established that AC = PC and AC = PB, so triangle APC and triangle APB are isosceles triangles with equal base angles.
9. Angle PAC = angle APC (both are base angles of isosceles triangle APC)
10. Angle PAB = angle APB (both are base angles of isosceles triangle APB)
Using these equal base angles, we can derive the following:
11. Angle BAC + angle PCA + angle PAC = 180 degrees (sum of angles in triangle APC)
12. Angle BAC + angle PAB + angle APB = 180 degrees (sum of angles in triangle APB)
Adding equations 11 and 12, we get:
13. 2 * (angle BAC + angle PAC) + 2 * angle PAB = 360 degrees
Since angle BAC + angle PAC + angle PAB = 180 degrees (from equation 7), we can substitute this into equation 13:
14. 2 * 180 degrees + 2 * angle PAB = 360 degrees
Simplifying equation 14, we get:
15. 360 degrees + 2 * angle PAB = 360 degrees
When we subtract 360 degrees from both sides, we obtain:
16. 2 * angle PAB = 0 degrees
Dividing both sides by 2, we get:
17. angle PAB = 0 degrees
Since angle PAB is an angle in triangle APB, it cannot be equal to 0 degrees. This means that our assumption in equation 11 and 12 must be incorrect.
Therefore, angle BPC + angle BQC cannot be equal to 2 right angles.
Hence, the statement "angle BPC + angle BQC = 2 right angles" is not true in general.
Apologies for the confusion caused.