Question

QI and RI bisect exterior angles RQT and QRS. Prove that `∠QIR = 90^circ - 1/2 ∠P`.

Answer 1

Given: In a triangle △PQR, let QI and RI bisect the exterior angles ∠RQT and ∠QRS respectively.

Prove that: `∠QIR = 90^circ - 1/2 ∠P`

Step 1: Understand the figure

1. △PQR

2. Let I be the intersection of the bisectors of the exterior angles at Q and R

QI bisects exterior angle at Q  ...(∠RQT)

RI bisects exterior angle at R  ...(∠QRS)

3. ∠QIR is the angle formed at I by these bisectors.

Step 2: Recall properties of exterior angle bisectors

The exterior angle at a vertex of a triangle:

Exterior angle = 180° – Interior angle

If a line bisects an exterior angle, it forms an angle with the side of the triangle that is:

`1/2 ("Exterior angle")`

= `1/2 (180^circ - "Interior angle")` 

= `90^circ - 1/2 ("Interior angle")`

Step 3: Apply to vertices Q and R

Let ∠Q = ∠PQR

Let ∠R = ∠PRQ

Let ∠P = ∠QPR

Exterior angles:

1. At Q: ∠RQT = 180° – ∠Q

2. At R: ∠QRS = 180° – ∠R

Their bisectors:

1. `∠IQX = 1/2 (180^circ - ∠Q) = 90^circ - 1/2 ∠Q`

2. `∠IRY = 1/2 (180^circ - ∠R) = 90^circ - 1/2 ∠R`

Step 4: Consider triangle formed by intersection I

At I, the angle ∠QIR is opposite angle P of triangle PQR because of the triangle’s angle sum property.

Property: In a triangle, if two lines bisect the exterior angles at two vertices, the angle formed at their intersection is:

`∠QIR = 180^circ - (90^circ + 1/2∠Q) - (90^circ + 1/2∠R)`

Angles between the bisectors at I:

∠QIR = 180° – (∠IQX + ∠IRY)

`∠QIR = 180^circ - (90^circ - 1/2∠Q) + (90^circ - 1/2∠R)`

`∠QIR = 180^circ - (180^circ - 1/2∠Q - 1/2∠R)`

`∠QIR = 1/2∠Q + 1/2∠R`

Step 5: Use triangle angle sum

∠P + ∠Q + ∠R = 180°

∠Q + ∠R = 180° – ∠P

`∠QIR = 1/2(180^circ - ∠P)`

`∠QIR = 90^circ - 1/2∠P`