# Ex.10.2 Q8 Circles Solution - NCERT Maths Class 10

## 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,

\(\begin{align}BP + CR + DR + AP \\= BQ + CQ + DS + AS\end{align}\)

On re-grouping them,

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

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.