#### Question

If the sides of a quadrilateral ABCD touch a circle, prove that:

AB + CD = BC + AD

#### Solution

Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.

Since AP and AS are tangents to the circle from external point A

AP = AS .......(i)

Similarly, we can prove that:

BP = BQ .......(ii)

CR = CQ .......(iii)

DR = DS ........(iv)

Adding,

AP + BP + CR + DR = AS + DS + BQ + CQ

AB + CD = AD + BC

Hence, AB + CD = AD + BC

Is there an error in this question or solution?

Solution If the Sides of a Quadrilateral Abcd Touch a Circle, Prove That: Ab + Cd = Bc + Ad Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.