Video: CBSE Class X • Pack 3 • 2016 • Question 7

CBSE Class X • Pack 3 • 2016 • Question 7

02:42

Video Transcript

In the given figure, a quadrilateral 𝐴𝐵𝐶𝐷 is drawn to circumscribe a circle, with centre 𝑂, such that the sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, and 𝐷𝐴 touch the circle at the points 𝑃, 𝑄, 𝑅, and 𝑆 respectively. Prove that 𝐴𝐵 plus 𝐶𝐷 is equal to 𝐵𝐶 plus 𝐷𝐴.

In order to answer this question, we need to know that the sides of a polygon that circumscribe a circle are tangents to that very circle. In this example, the line 𝐵𝐶 is a tangent to the circle at the point 𝑄. And the line 𝐶𝐷 is also a tangent to the circle at the point 𝑅. We know that two tangents that meet at a point are of equal lengths. This means that the line segments joining the points 𝐴 and 𝑃 and the points 𝐴 and 𝑆 are of equal lengths.

Similarly, the line segments that join 𝐵 to 𝑃 and 𝐵 to 𝑄 are also of equal lengths. We can create two more equations of this form. 𝐶𝑅 is equal to 𝐶𝑄 and 𝐷𝑅 is equal to 𝐷𝑆. This next step is going to seem a little bit strange. We’re going to add all of these four equations. Adding the left-hand sides, and we get 𝐴𝑃 plus 𝐵𝑃 plus 𝐶𝑅 plus 𝐷𝑅. And adding the right-hand sides, we can see that this is equal to 𝐴𝑆 plus 𝐵𝑄 plus 𝐶𝑄 plus 𝐷𝑆.

Looking at our diagram though, we can see that the line 𝐴𝐵 is made up of the line segments 𝐴𝑃 and 𝐵𝑃. So 𝐴𝑃 plus 𝐵𝑃 must be equal to 𝐴𝐵. Similarly, the line that joins 𝐶 to 𝐷 is made up of the two line segments 𝐶𝑅 and 𝐷𝑅. So that means 𝐶𝑅 plus 𝐷𝑅 is equal to the line 𝐶𝐷. Let’s continue this pattern on the right-hand side of the equation. The line 𝐵𝐶 is made up of the two line segments joining 𝐵 and 𝑄 and 𝐶 and 𝑄. This means that 𝐵𝑄 plus 𝐶𝑄 is equal to 𝐵𝐶.

Finally, we can see that the line joining 𝐷 to 𝐴 is made up of the two line segments joining 𝐴 and 𝑆 and 𝐷 and 𝑆. So 𝐴𝑆 plus 𝐷𝑆 is equal 𝐷𝐴. And that’s it. We’ve done it! We have proven that 𝐴𝐵 plus 𝐶𝐷 is equal to 𝐵𝐶 plus 𝐷𝐴 in this scenario.

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