Question

A quadrilateral ABCD is draw to circumscribe circle if AB =12,BC=15 ,and CD = 14 the AD is equal to ______.​

Answer 1

Circles.

Answer :

Proved!

Step-by-step explanation :

Given that :

ABCD be the quadrilateral circumscribe a circle with O.

The quadrilateral touches the circle at points P, Q, R & S.

To Prove :

AB + CD = AD + BC.

Solution :

We know that :

$\bigstar\;\textbf{\underline{\underline{Lengths of the tagnets drawn from external point are Equal.}}}$

So,

$\begin{gathered} \\ \\\implies \sf{AP = AS......Eq(1)}\\ \\\implies \sf BP = BQ.......Eq(2)\\ \\\implies \sf{CR = CQ.......Eq(3)}\\ \\\implies \sf{DR = DS..........Eq(4)}\end{gathered} </p><p>$

$\textbf{\underline{\underline{Addition of all Equations,}}}$

$\sf AP + BP+CR+DR = AS =BQ+CQ=DS \\ \\\implies \sf(AP+BP) + (CR+DR)=(AS+SD) +(BQ+CQ)\\ \\\implies \sf AB+CD = AD + BC$

$\textbf{\underline{\underline{Hence, Proved!}}}$

Refer to the Attachment for the figure.

Answer 2

Answer:

Circles.

Answer :

Proved!

Step-by-step explanation :

Given that :

ABCD be the quadrilateral circumscribe a circle with O.

The quadrilateral touches the circle at points P, Q, R & S.

To Prove :

AB + CD = AD + BC.

Solution :

We know that :

\bigstar\;\textbf{\underline{\underline{Lengths of the tagnets drawn from external point are Equal.}}}★

Lengths of the tagnets drawn from external point are Equal.

So,

\begin{gathered}\begin{gathered} \\ \\\implies \sf{AP = AS......Eq(1)}\\ \\\implies \sf BP = BQ.......Eq(2)\\ \\\implies \sf{CR = CQ.......Eq(3)}\\ \\\implies \sf{DR = DS..........Eq(4)}\end{gathered}\end{gathered}

⟹AP=AS......Eq(1)

⟹BP=BQ.......Eq(2)

⟹CR=CQ.......Eq(3)

⟹DR=DS..........Eq(4)

Addition of all Equations,

AP + BP+CR+DR = AS =BQ+CQ=DS (AP+BP) + (CR+DR)=(AS+SD) +(BQ+CQ)\AB+CD = AD + BC

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

⟹(AP+BP)+(CR+DR)=(AS+SD)+(BQ+CQ)

⟹ABHence, Proved!

Hence, Proved!

Refer to the Attachment for the figure.