Question

Seg AB is the diameter of a circle with centre P. Line l is a tangent to the circle at M.

Tangents to the circle at A and B intersect line l in the points C and D. Show that

a) Line AC || line BD

b) m∠CPD = 90°​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

m∠CPD = 90°​

Step-by-step explanation:

AP = PB = PM _____________(1) (Radii of a same circle)

AC = MC & BD = MD _______(2) (Tangent segment theorem)

∠CAP = ∠DBP = 90°​ ________(3) (tangent theorem)

Construction: Join seg AM & seg BM.

In Quadrilateral APMC,

AC ≅ CM ___________ from (2)

AP ≅ PM ___________ from (1)

Thus quadrilateral APMC is a Kite

Hence, seg PC ⊥ seg AM at point S _______(4) (diagonal of a kite are perpendicular to each other)

∴ ∠PSM = 90° ___________(5)

In  Quadrilateral BPMD,

BD ≅ DM ___________ from (2)

BP ≅ PM ____________ from (1)

Thus quadrilateral BPMD is a Kite

Hence, seg PD ⊥ seg BM at point T ___________(6) (diagonal of a kite are perpendicular to each other)

∴ ∠PTM = 90° ___________(7)

∠AMB = 90° ____________(8) (Angle inscribed in semicircle)

In Quadrilateral PSMT,

∠S = ∠M = ∠T = 90° _________[from (5), (7) and (8)]

∴ ∠SPT = 90° _____________ (By angle sum property of a quadrilateral)

∴ ∠CPD = 90° _____________(C-S-P, D-T-P)

HENCE PROVED !