Adv
AnilSinghBora
  • 0
Guru

Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre. Q.4

  • 0

How can i solve this tough question of class 9th ncert of Circles of math of exercise 10.6 of question no.4 . Give me the best and simple way for solving this question. Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Share

1 Answer

  1. Consider the diagram

    Ncert solutions class 9 chapter 10-35

    Here AD = CE

    We know, any exterior angle of a triangle is equal to the sum of interior opposite angles.

    So,

    ∠DAE = ∠ABC+∠AEC (in ΔBAE) ——————-(i)

    DE subtends ∠DOE at the centre and ∠DAE in the remaining part of the circle.

    So,

    ∠DAE = (½)∠DOE ——————-(ii)

    Similarly, ∠AEC = (½)∠AOC  ——————-(iii)

    Now, from equation (i), (ii), and (iii) we get,

    (½)∠DOE = ∠ABC+(½)∠AOC

    Or, ∠ABC = (½)[∠DOE-∠AOC]  (hence proved).

    • 0
Leave an answer

Leave an answer

Browse

Choose from here the video type.

Put Video ID here: https://www.youtube.com/watch?v=sdUUx5FdySs Ex: "sdUUx5FdySs".

Captcha Click on image to update the captcha.

Related Questions