Prove that the angle subtended by an Arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
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This can be proved as follows :
Given a circle (C,r) in which arc AB subtends angle (ACB) at the centre and angle (ADB) at any point D on the circle.
To Prove:
(i) angle (ACB) = 2×angle (ADB), when arc AB is a minor arc or a semicircle.
(ii) Reflex angle (ACB)=2× angle (ADB), when arc AB is a major arc
Construction: Join AB and DC. Produce DC to a point X outside the circle.
Clearly, there are 3 cases.
Case I : arc AB is a minor arc [Fig (i)]
Case II : arc AB is a semicircle [Fig (ii)]
Case III : arc AB is a major arc [Fig (iii)]
We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.
Therefore,
angle (ACX) = angle (CAD) + angle (CDA) [consider triangle CAD]
angle (BCX) = angle (CBD) + angle (CDB) [consider triangle CBD]
But,
angle (CAD) = angle (CDA) [since CD=CA=r ] and ,
angle (CBD) = angle (CDB) [since CD=CB=r ]
Therefore,
angle (ACX)=2× angle (CDA) and
angle (BCX)=2×angle (CDB)
In Fig (i) and Fig (ii) ,
angle (ACX) + angle (BCX) = 2× angle (CDA) + 2× angle (CDB)⇒
angle (ACB) = 2[angle (CDA) + angle (CDB)] ⇒
angle (ACB) = 2×angle (ADB)
in Fig (iii),
angle (ACX) + angle (BCX) = 2× angle (CDA) + 2× angle (CDB)⇒
reflex angle (ACB) = 2[angle (CDA) + angle (CDB)] ⇒
reflex angle (ACB) = 2× angle (ADB)
Hence proved.
This theorem has 2 main implications:
(i) The angle in a semicircle is a right angle .
i.e. in fig (2), where arc AB is a semicircle, or in other words ACB is a diameter,
angle (ADB)=12× angle (ACB)=12×(180 degrees) = 90 degrees = a right angle
This holds true for any position of point C on the semicircle.
(ii) Angles in the same segment of a circle are equal.
This means that if an arc subtends 2 angles, at 2 different points on the circle, these angles will be equal.
We can prove this, by proving that each of the 2 angles is equal to half the angle subtended at the centre.
Given a circle (C,r) in which arc AB subtends angle (ACB) at the centre and angle (ADB) at any point D on the circle.
To Prove:
(i) angle (ACB) = 2×angle (ADB), when arc AB is a minor arc or a semicircle.
(ii) Reflex angle (ACB)=2× angle (ADB), when arc AB is a major arc
Construction: Join AB and DC. Produce DC to a point X outside the circle.
Clearly, there are 3 cases.
Case I : arc AB is a minor arc [Fig (i)]
Case II : arc AB is a semicircle [Fig (ii)]
Case III : arc AB is a major arc [Fig (iii)]
We know that when one side of a triangle is produced then the exterior angle so formed is equal to the sum of the interior opposite angles.
Therefore,
angle (ACX) = angle (CAD) + angle (CDA) [consider triangle CAD]
angle (BCX) = angle (CBD) + angle (CDB) [consider triangle CBD]
But,
angle (CAD) = angle (CDA) [since CD=CA=r ] and ,
angle (CBD) = angle (CDB) [since CD=CB=r ]
Therefore,
angle (ACX)=2× angle (CDA) and
angle (BCX)=2×angle (CDB)
In Fig (i) and Fig (ii) ,
angle (ACX) + angle (BCX) = 2× angle (CDA) + 2× angle (CDB)⇒
angle (ACB) = 2[angle (CDA) + angle (CDB)] ⇒
angle (ACB) = 2×angle (ADB)
in Fig (iii),
angle (ACX) + angle (BCX) = 2× angle (CDA) + 2× angle (CDB)⇒
reflex angle (ACB) = 2[angle (CDA) + angle (CDB)] ⇒
reflex angle (ACB) = 2× angle (ADB)
Hence proved.
This theorem has 2 main implications:
(i) The angle in a semicircle is a right angle .
i.e. in fig (2), where arc AB is a semicircle, or in other words ACB is a diameter,
angle (ADB)=12× angle (ACB)=12×(180 degrees) = 90 degrees = a right angle
This holds true for any position of point C on the semicircle.
(ii) Angles in the same segment of a circle are equal.
This means that if an arc subtends 2 angles, at 2 different points on the circle, these angles will be equal.
We can prove this, by proving that each of the 2 angles is equal to half the angle subtended at the centre.
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