CBSE · Class 10 · Mathematics · CirclesA quadrilateral $ABCD$ is drawn to circumscribe a circle. Prove that $AB + CD = AD + BC$. Further, if a circle touches all the four sides of a parallelogram $ABCD$, prove that the parallelogram is a rhombus.

Part 1: Prove that $AB + CD = AD + BC$

Theorem Statement & Proof:\nWe know that the lengths of tangents drawn from an external point to a circle are equal. \nLet the circle touch the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$ respectively.

1. From external point $A$: $$AP = AS \quad \text{--- (1)}$$
2. From external point $B$: $$BP = BQ \quad \text{--- (2)}$$
3. From external point $C$: $$CR = CQ \quad \text{--- (3)}$$
4. From external point $D$: $$DR = DS \quad \text{--- (4)}$$ \nAdding equations (1), (2), (3), and (4): $$(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$$ \nObserving the segments from the figure:

• $AP + BP = AB$
• $CR + DR = CD$
• $AS + DS = AD$
• $BQ + CQ = BC$ \nSubstituting these into the sum equation: $$AB + CD = AD + BC$$ (Hence Part 1 proved)

Part 2: Prove that parallelogram $ABCD$ is a rhombus

Given:

• $ABCD$ is a parallelogram.
• A circle touches all four sides of $ABCD$.

To Prove:

• $ABCD$ is a rhombus (i.e., all four sides are equal: $AB = BC = CD = DA$).

Proof:

1. Since $ABCD$ is a parallelogram, its opposite sides are equal:

   • $AB = CD \quad \text{--- (A)}$
   • $AD = BC \quad \text{--- (B)}$

2. From Part 1, we already established that for any circumscribing quadrilateral: $$AB + CD = AD + BC$$

3. Substitute equations (A) and (B) into this relation:

   • Replace $CD$ with $AB$ and $BC$ with $AD$: $$AB + AB = AD + AD$$ $$2AB = 2AD$$ $$AB = AD \quad \text{--- (C)}$|

4. Combining equation (A) and (C):

   • Since $AB = CD$ and $AB = AD$, we have $AB = BC = CD = DA$ (because $AD = BC$).

5. Conclusion:

   • A parallelogram with all equal sides is defined as a rhombus.
   • Therefore, $ABCD$ is a rhombus.