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
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 = \(\frac{1}{2}\)∠AOC ......(iii)
From equations (ii) and (iii),
∠ABC = ∠AOE ......(iv)
Now, in ΔOEP
AE is the bisector of ∠OEP
\(\frac{OA}{AP} = \frac{OE}{EP}\)
⇒ 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.
