Question

Seg AB is a diameter of a circle woth centre C. Line PQ is a tangent, which touches the circle at pt T. Seg AP perpendicular to line PQ and BQ perpendicular to line PQ. Prove: CP=CQ​

Answer 1

Step-by-step explanation:

since PQ is a tangent to the circle, ∠CTP = ∠CTQ = 90

Since ∠APT = ∠CTP = 90

AP || CT.

Similarly CT || BQ.

From association, we can say AP || CT || BQ

AB is a line cutting all three parallel lines.

AC = CB (Radius of the circle, and AB is diameter, C is center)

Since C is center point of line AB cutting parallel lines.

We can say these parallel lines are equal distance.

Hence PT = TQ.

Now in ΔCTP and ΔCTQ,

CT is a common side, PT = TQ and ∠CTP = ∠CTQ = 90

So we can prove CP = CQ. (Pythagoras theorem or congruent triangle theorem)

Answer 2

Answer:

Proved

Step-by-step explanation:

Seg AB is a diameter of a circle woth centre C. Line PQ is a tangent, which touches the circle at pt T. Seg AP perpendicular to line PQ and BQ perpendicular to line PQ. Prove: CP=CQ​

In Δ ACP

CP/ Sin∠A =  AC/Sin∠x

where ∠x = ∠CPA    ∠A=∠PAC

=> AC Sin∠A = CP Sin∠x   - eq 1

In Δ BCQ

CQ/Sin∠B = BC/Sin∠y

∠y = ∠CQB    ∠A=∠CBQ

∠A + ∠B + 90° + 90° = 360°  ( angles of quadrilateral ABQP)

=> ∠A + ∠B = 180°°

∠B = 180° - ∠A

Sin∠B = Sin(180° - ∠A) = SinA

CQ/Sin∠A = BC/Sin∠y

=> BC Sin∠A = CQ Sin∠y

BC = AC + radius

=> AC Sin∠A = CQ Sin∠y   - eq 2

equating eq 1 & Eq 2

CP Sin∠x = CQ Sin∠y

Squaring both sides

CP² Sin²∠x = CQ² Sin²∠y    - Eq 3

in ΔCPT

Sin(90-x) = CT/CP

Cosx = CT/CP

=> CT = CPCosx    - Eq 4

in ΔCQT

Sin(90-y) = CT/CQ

Cosy = CT/CQ

=> CT = CPCosy   - Eq 5

Equating eq 4 & 5

CPCosx = CPCosy

Squaring both sides

CP² Cos²∠x = CQ² Cos²∠y   - Eq 6

Adding Eq 3 & Eq 6

CP² Sin²∠x + CP² Cos²∠x = CQ² Sin²∠y + CQ² Cos²∠y

=> CP²(Sin²∠x + Cos²∠x ) = CQ²(Sin²∠y + Cos²∠y )

Sin²θ + Cos²θ = 1

=> CP² = CQ²

=> CP = CQ

QED