a circle is touching the sides BC of /_\ ABC at P.AB and AC when produced are touching the circle at Q and R respectively, Prove that AQ=1/2 (AB+BC+CA)
To prove that AQ = 1/2 (AB + BC + CA), we can use the concept of tangents from a point to a circle.
First, let's draw the given figure:
```
C
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Q--------P
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|/
R
```
In triangle ABC, we have a circle tangent to side BC at point P. We are given that AB and AC, when produced, are touching the circle at points Q and R, respectively.
To prove AQ = 1/2 (AB + BC + CA), we will divide the proof into two parts:
Part 1: Prove AQ = AP
Part 2: Prove AP = 1/2 (AB + BC + CA)
Part 1: Prove AQ = AP
Since AP is a tangent to the circle, we can conclude that angle AQP = 90 degrees (tangent meets radius at 90 degrees).
In triangle AQP, we have angle AQP = 90 degrees, which means triangle AQP is a right triangle. By the definition of a right triangle, the hypotenuse (AQ) is equal to the side opposite the right angle. Therefore, we can conclude that AQ = AP.
Part 2: Prove AP = 1/2 (AB + BC + CA)
Now, let's consider triangle ABC. From the given information, we know that the circle is tangent to side BC at point P. This means that AP is a common tangent to the circle.
Since a tangent to a circle is perpendicular to the radius at the point of tangency, we can conclude that angle APC = 90 degrees.
Now, let's consider the points where AB and AC (produced) touch the circle, which are Q and R, respectively.
In triangle ABC, angle ACB is equal to the sum of angles QCB (exterior angle) and QBC (interior angle). Similarly, angle ABC is equal to the sum of angles RBC (exterior angle) and RCB (interior angle).
Therefore, angle ACB = QCB + QBC
= RBC + RCB
If we substitute these angle relationships in triangle APC, we get:
angle APC = angle ACB
= QCB + QBC
= RBC + RCB
Since angle APC is common to both sides of the equation, we can conclude that QBC = RBC and QCB = RCB.
Therefore, QB = RB and QC = RC.
By the tangent-chord theorem, we know that the lengths of the tangents drawn from an external point to a circle are equal. This means that QB = QC.
Now, let's consider triangle AQD, where D is the midpoint of BC.
In triangle AQD, we have AD = 1/2 BC (by the definition of midpoint).
Since QB = QC and BD = DC (because D is the midpoint of BC), we can apply the triangle inequality theorem to triangle AQD:
AQ + AD > QD
AQ + 1/2 BC > QD
Rearranging the inequality, we have:
AQ > QD - 1/2 BC
Since QD is the length of the common tangent from Q to the circle, QD = QB = QC.
Therefore, we get:
AQ > QB - 1/2 BC
AQ > QB - BC/2
Substituting QB = QC, we have:
AQ > QC - BC/2
AQ > QC - QC/2
AQ > QC/2
But we also have AQ = AP from Part 1 of the proof.
So, we can write:
AP > QC/2
Since AP is a tangent to the circle, it is equal to the radius of the circle.
Let's denote the radius of the circle as 'r'.
Therefore, AP = r.
Substituting this in the inequality, we get:
r > QC/2
Now, let's consider triangle ABC again.
If we add up all the sides, we get:
AB + BC + CA = QC + CQ + QB + BA + AC
Since QB = QC, we can simplify this as:
AB + BC + CA = 2QC + BA + AC
Now divide both sides by 2, we have:
(AB + BC + CA) / 2 = QC + 1/2 (BA + AC)
Since QC is the radius of the circle, we can substitute r for QC:
(AB + BC + CA) / 2 = r + 1/2 (BA + AC)
Now, we know that r > QC/2, so we can substitute r with QC/2:
(AB + BC + CA) / 2 = QC/2 + 1/2 (BA + AC)
Finally, we have:
(AB + BC + CA) / 2 = AQ
Hence, AQ = 1/2 (AB + BC + CA).
Therefore, we have proved the desired result.