**Q) **In figure 1, a right triangle ABC in which ∠B = 90 is shown. Taking AB as diameter, a circle has been drawn intersecting AC at point P. Prove that the tangent drawn at point P bisects BC.

**Ans: Here is the step by step solution method:**

**Step 1:**

We can see that, Q is an external point to the circle and PQ and BQ are the tangents drawn on a circle

∴ QB = QP………….(i)

Let’s connect Point P and B:

Now since, in a triangle, if the sides are equal, their opposite angles will also be equal,

∴ ∠QBP = ∠BPQ …………… (ii)

**Step 2:**

Next, since AB is the diameter of the circle and teh angle made by diameter on a cirle is 90,

∴ ∠APB = 90

Next, On point P we have a pair of angles

∴ ∠APB + ∠CPB = 180

∠CPB = 180 − 90 = 90 ……….(iii)

Next, we can see that Line PQ is dividing ∠CPB

∴ ∠CPQ + ∠BPQ = 90 ………..(iv)

**Step 3:**

Further in △CPB, by Angle sum property, we get:

∠CPB + ∠CBP + ∠BCP = 180

Substituting value of ∠CPB from equation (iii), we get:

90 + ∠CBP + ∠BCP = 180

∴ ∠CBP + ∠BCP = 180 − 90 = 90 ………..(v)

**Step 4:**

Now by comparing equations (iv) and (v), we get

∠CBP + ∠BCP = ∠CPQ + ∠BPQ ………… (vi)

Since, ∠CBP can be written as ∠QBP and ∠BCP can be written as ∠QCP, therefore equation (vi) can be written as:

∠QBP + ∠QCP = ∠CPQ + ∠BPQ ………… (vii)

Next, we substitute value of ∠QBP from equation (ii) in equation (vi), we get:

(∠BPQ) + ∠QCP = ∠CPQ + ∠BPQ

∴ ∠QCP = ∠CPQ ………… (viii)

**Step 5:**

Next in △PQC, if the angles are equal, their opposite sides will also be equal,

here, since ∠QCP = ∠CPQ

∴ QP = QC ………..(ix)

Now by comparing equations From (i) and (ix), we get

QB = QP = QC

∵ QB = QC,

Therefore, tangent to the circle PQ, bisects the side BC.

**Hence Proved !**

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