Question

2. Prove that the parallelogram ABCD circumscribing a circle with center E is a
morbus
​

Answer 1

Answer:

ABCD is a Rhombus

Step-by-step explanation:

A circle with center E

A parallelogram Touching the circle at points PQR and S

To prove ‐

ABCD is a Rhombus

Proof

A Rhombus Is a parallelogram with all sides equal

so we have to prove all sides equal

In parallelogram ABCD

AB=CD & AD=BC

Hence

AP=AS

BP=BQ

CR=CQ

DR=DS

adding 2+3+4+5

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

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

AB+CD=AD+BC

AB+AB=AD+AD

2AD=2AB

AB=AD

AB=CD & AD=CD

So ABCD is a parallelogram

ABCD is a Rhombus

Hence proved