Notifications
Clear all
Circle
1
Posts
2
Users
0
Likes
169
Views
0
20/06/2021 11:13 am
Topic starter
Prove that the parallelogram circumscribing a circle is a rhombus.
1 Answer
0
20/06/2021 11:15 am
Now, since ABCD is a parallelogram, AB = CD and BC = AD.
From the above figure, it is seen that,
(i) DR = DS
(ii) BP = BQ
(iii) CR = CQ
(iv) AP = AS
These are the tangents to the circle at D, B, C, and A respectively.
Adding all these we get,
DR+BP+CR+AP = DS+BQ+CQ+AS
By rearranging them we get,
(BP+AP)+(DR+CR) = (CQ+BQ)+(DS+AS)
Again by rearranging them we get,
AB+CD = BC+AD
Now, since AB = CD and BC = AD, the above equation becomes
2AB = 2BC
∴ AB = BC
Since AB = BC = CD = DA, it can be said that ABCD is a rhombus.