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Important questions based on NCERT syllabus for Chapter 10 - Circles:
Question 1: In Fig., if AB = AC, prove that BE = EC
Since tangents from an exterior point to a circle are equal in length
Therefore, AD = AF ... (i) [Tangents from A]
BD = BE ....(ii) [Tangents from B]
CE = CF ....(iii) [Tangents from C]
Now, AB = AC
⇒ AB – AD = AC – AD
[Subtracting AD from both sides]
⇒ AB – AD = AC – AF [Using (i)]
⇒ BD = CF
⇒ BE = CF [Using (ii)]
⇒ BE = CE [Using (iii)]
Question 2: PA and PB are tangents from P to the circle with center O. At point M which lies on circle, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN.
We know that the tangents drawn from an external point to a circle are equal in length.
KA = KM .... (i) [From K]
and, NB = NM .... (ii) [From N]
Adding equations (i) and (ii), we get
KA + NB = KM + NM
⇒ AK + BN = KM + MN
⇒ AK + BN = KN
Question 3: Prove that the tangents at the extremities of any chord make equal angles with the chord.
Let AB be a chord of a circle with center O, and let AP and BP be the tangents at A and B respectively.
Suppose the tangents meet at P .
Join OP. Suppose OP meets AB at C . We have to prove that ∠PAC = ∠PBC .
In triangles PCA and PCB , We have
PA = PB
∠APC = ∠BPC
[The tangents are equally inclined to line joining extrernal point and centre of circle.]
PC = PC [Common]
So, by SAS - criterion of congruence, we have
ΔPAC ≅ ΔPBC
⇒ ∠PAC = ∠PBC