(a) Construct a triangle ABC such that BC = 8 cm, AC = 10 cm and ∠ABC = 90°.

(b) Construct an incircle to this triangle. Mark the centre as I.

(c) Measure and write the length of the in-radius.

(d) Measure and write the length of the tangents from vertex C to the incircle.

(e) Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively. Write the relationship between ∠RIQ and ∠QCR.

(Use a ruler and a compass for this question.)

Steps of construction :

1. Draw a line segment BC = 8 cm.

2. Draw BX perpendicular to BC.

3. With C as center and radius = 10 cm, draw an arc cutting BX at A.

4. Join AB and AC.

5. Draw AW, BY and CZ the angle bisectors of A, B and C respectively.

6. Mark the point of intersection as I.

7. Draw IR perpendicular to side AC.

8. With I as center and radius IR draw a circle, which is the required incircle.

9. Mark points P, Q and R where the incircle touches the sides AB, BC, and AC of the triangle respectively.

10. Measure CQ and CR.

From figure,

⇒ ∠IRC = ∠IQC = 90° (The radius from the center of the circle to the point of tangency is perpendicular to the tangent line.)

⇒ ∠RCI = ∠QCI = $\dfrac{∠C}{2}$ (As CZ is angle bisector)

In △ IRC,

⇒ ∠RIC = 180° - ∠RCI - ∠IRC [∵ Sum of ∠'s in a Δ = 180°]

⇒ ∠RIC = 180° - $\dfrac{∠C}{2}$ - 90°

⇒ ∠RIC = 90° - $\dfrac{∠C}{2}$ …………(1)

In △ IQC,

⇒ ∠QIC = 180° - ∠IQC - ∠ICQ [∵ Sum of ∠'s in a Δ = 180°]

⇒ ∠QIC = 180° - 90° - $\dfrac{∠C}{2}$

⇒ ∠QIC = 90° - $\dfrac{∠C}{2}$ …………(2)

Adding equations (1) and (2), we get :

⇒ ∠RIC + ∠QIC = 90° - $\dfrac{∠C}{2}$ + 90° - $\dfrac{∠C}{2}$

⇒ ∠RIQ = 180° - ∠C

⇒ ∠RIQ = 180° - ∠RCQ

⇒ ∠RIQ + ∠RCQ = 180°.

Hence, ∠RIQ + ∠QCR = 180°.