- Let the vertex of an angle ABC be located outside a circle and let t

1.	- Let the vertex of an angle ABC be located outside a circle and let the sidesof the angle intersect equal chords AD and CE with the circle. Prove thatABC is equal to half the difference of the angles subtended by the chordsAC and DE at the centre.
Answer» Given:AD = CE To prove:∠ABC = 1/2(∠DOE –∠AOC) In△AOD and△COE AD = CE (Given) AO = OC and DO = OE (Radii of same circle) ∴△AOD ≅ △COE (By SSS congruence criterion) ⇒∠1 =∠3,∠2 =∠4 (CPCT) ....(i) But OA = OD and OC = OE ⇒∠1 =∠2 and∠3 =∠4 .....(ii) From (i) and (ii), we have ∠1 =∠2 =∠3 =∠4 (= x say) Also, OA = OC and OD = DE ⇒∠7 =∠8 (= z say) and∠5 =∠6 (= y say) Now, ADEC is a cyclic quadrilateral ⇒∠DAC +∠DEC =180° ⇒ x + z + x + y =180° ⇒ y =180°- 2x - z ...(iii) In△DOE,∠DOE = 180° - 2y and in△AOC,∠AOC = 180° - 2z ∴∠DOE -∠AOC = (180° - 2y) - (180° - 2z) = 2z - 2y = 2z - 2(180° - 2x - z) (Using (iii)) = 4z + 4x - 360° ....(iv) Again, ∠BAC + ∠CAD = 180° ⇒∠BAC = 180°– (z + x) ....(v) Similarly, ∠BCA = 180°– (z + x) ....(vi) In△ABC,∠ABC =180° -∠BAC - ∠BCA =180° - 2[180° - (z - x)] (Using (v) and (vi)) = 2z + 2x -180° = 1/2(4z + 4x - 360°) ....(vii) From (iv) and (vii), we have ∠BAC = 1/2(∠DOE - ∠AOC)

- Let the vertex of an angle ABC be located outside a circle and let the sidesof the angle intersect equal chords AD and CE with the circle. Prove thatABC is equal to half the difference of the angles subtended by the chordsAC and DE at the centre.

Answer»

Given:AD = CE

To prove:∠ABC = 1/2(∠DOE –∠AOC)

In△AOD and△COE

AD = CE (Given)

AO = OC and DO = OE (Radii of same circle)

∴△AOD ≅ △COE (By SSS congruence criterion)

⇒∠1 =∠3,∠2 =∠4 (CPCT) ....(i)

But OA = OD and OC = OE ⇒∠1 =∠2 and∠3 =∠4 .....(ii)

From (i) and (ii), we have

∠1 =∠2 =∠3 =∠4 (= x say)

Also, OA = OC and OD = DE

⇒∠7 =∠8 (= z say) and∠5 =∠6 (= y say)

Now, ADEC is a cyclic quadrilateral

⇒∠DAC +∠DEC =180°

⇒ x + z + x + y =180° ⇒ y =180°- 2x - z ...(iii)

In△DOE,∠DOE = 180° - 2y

and in△AOC,∠AOC = 180° - 2z

∴∠DOE -∠AOC = (180° - 2y) - (180° - 2z) = 2z - 2y

= 2z - 2(180° - 2x - z) (Using (iii))

= 4z + 4x - 360° ....(iv)

Again, ∠BAC + ∠CAD = 180° ⇒∠BAC = 180°– (z + x) ....(v)

Similarly, ∠BCA = 180°– (z + x) ....(vi)

In△ABC,∠ABC =180° -∠BAC - ∠BCA

=180° - 2[180° - (z - x)] (Using (v) and (vi))

= 2z + 2x -180° = 1/2(4z + 4x - 360°) ....(vii)

From (iv) and (vii), we have

∠BAC = 1/2(∠DOE - ∠AOC)