Ex.10.2 Q8 Circles Solution - NCERT Maths Class 10

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Question

A quadrilateral \(ABCD\) is drawn to circumscribe a circle (see given Figure). Prove that: \(AB + CD = AD + BC\)

Text Solution

 

To prove:

\(AB + CD = AD + BC\)

Reasoning:

  • A tangent to a circle is a line that intersects the circle at only one point.
  • Theorem 10.1: The tangent at any point of a circle is perpendicular to the radius through the point of contact.
  • Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

Steps:

We know that the length of tangents drawn from an external point of the circle are equal according to Theorem \(10.2.\)

Therefore,

\(BP = BQ\) (Tangents from point \(B\)) \(\dots\dots\rm{(i)}\)

\(CR = CQ\) (Tangents from point \(C\)) \(\dots \dots\rm{(ii)}\)

\(DR = DS\)  (Tangents from point \(D\)) \(\dots \dots\rm{(iii)}\)

\(AP = AS\) (Tangents from point \(A\)\(\dots \dots\rm{(iv)}\)

Adding all the four equations, \(\text{( i) +( ii) + (iii )+ (iv)}\) we get,

\(BP + CR + DR + AP = BQ + CQ + DS + AS\)

On re-grouping them,

\(( A P + B P ) + ( C R + D R ) = ( A S + D S ) + ( B Q + C Q )\, .... \,( \rm{v} )\)

From the figure, we can see that:

\[\begin{align} {AP + BP} &= {AB} \\ {CR + DR} &= {CD} \\{AS + DS} &= {A D}\\{BQ + CQ} &= {B C}\end{align}\]

On substituting the above values in \(\rm{(v),}\) we get

\(AB + CD = AD + BC\)

Hence Proved.

  
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