Question

Given: O is the centre of the circle, AB is a diameter, OA = AP, O – A – P, PC is a tangent through C. A tangent through point A intersects PC in E and BC in D.

To prove: ΔCED is an equilateral triangle.

Answer 1

Given: OA = AP, PC is a tangent at C, AD is tangent at A.

To prove: ΔCED is an equilateral angle.

Construction: Join OC and OE.

Proof: AD is a tangent to the circle at A.

∴ ∠ACP = ∠ABC  ......(i) [Alternate segment theorem]

Since the measure of the angle an arc subtends at a point on a circle is half that of the angle the arc subtends at its centre

∴ ∠ABC = `1/2` ∠AOC ......(ii)

In ΔAOE and ΔCOE

OA ≅ OC  ......[Radii of circle]

OE ≅ OE  ......[Common side]

AE ≅ CE  ......[Tangents to the circle from point E]

∴ By SSS criterion of congruence,

ΔAOE ≅ ΔCOE

∴ ∠AOE = ∠COE  ......[C.A.C.T]

or ∠AOE = ∠COE  = `1/2` ∠AOC  ......(iii)

From equations (ii) and (iii),

∠ABC = ∠AOE  ......(iv)

Now, in ΔOEP

AE is the bisector of ∠OEP

∴ `(OA)/(AP) = (OE)/(EP)`

⇒ OE = EP  ......[∵ OA = AP, Given]

∴ ∠AOE = ∠APE  .....(v) [Angles opposite to equal sides]

or ∠ABC = ∠APE  .....(vi) [Using equation (iv)]

From the figure,

∠DCE = 90° – ∠ACP  .....(vii)

∠CDE = 90° – ∠ABC  ......(viii)

∠AEP = ∠CED = 90° – ∠APE  ......(ix) [∠AEP = ∠CED; Vertically opposite angles]

From equations (vii), (viii) and (ix)

∠DCE = ∠CDE = ∠CED

or DE = CE = CD  ......[Sides opposite to equal angles]

Hence ΔCED is an equilateral triangle.

Hence Proved.