# If the sides of a quadrilateral ABCD touch a circle, then: AB + BC = CD + AD AB + CD = BC + AD AB - CD = BC - AD AB - CD = BC + AD

## Question

If the sides of a quadrilateral ABCD touch a circle, then:

- AB + BC = CD + AD
- AB + CD = BC + AD
- AB - CD = BC - AD
- AB - CD = BC + AD

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## Answer

The correct option is:

$$ AB + CD = BC + AD $$

**Given:**
A circle touches the sides $AB$, $BC$, $CD$, and $DA$ of quadrilateral $ABCD$ at points $P$, $Q$, $R$, and $S$, respectively.

**To Prove:**
$$
AB + CD = BC + AD
$$

**Proof:**

**By Theorem:** Tangents drawn from an external point to a circle are equal in lengths.

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

$$ AP = AS $$

**Similarly:**

$$ BP = BQ \ CR = CQ \ DR = DS $$

Adding these equations:

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

This simplifies to:

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

Rewriting it, we have:

$$ AB + CD = AD + BC $$

Hence, we conclude:

$$ AB + CD = BC + AD $$

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