Question

if ABCD is a parallelogram the prove that

Answer 1

AB=AB(common)
AD=BC(opposite sides of ||gm are equal)
Angle A=Angle C(90°)
Thus triangle ABD=~ triangle BCD(S. A. S. axiom)

Answer 2

Answer:

Step-by-step explanation:

Given:

   A parallelogram ABCD circumscribing a circle.

To Prove:

   ABCD is a rhombus?

Solution: See the figure above , we have

   AB = DC {Opposite sires of ||gm}

   AD = BC {Opposite sires of ||gm}

As we know that length of two tangents drawn from a external points to a circle are equal to each other.

Let's see the tangents form each point

   AP = AS {from point A}.....i

   BP = BQ {from B}.....ii

   CR = CQ {from C}.....iii

   DR = DS {from D}.....iv

Now,

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

➯ AB + CD = BC + AD

➯ 2AB = 2AD (opposite sides are equal)

➯ AB = AD

∴ABCD is a ||gm with adjacent sides AB = AD. Hence, is a rhombus.