Question

A quadrilateral ABCD is drawn to circumscibe a circle . Proof that AB+CD=AD+BC

Answer 1

Step-by-step explanation:

$wkt \\ \\ the \: lenght \: of \: tangents \: drawn \\from \: an \: external \: point \: o \: the \: circle \: are \: equal \\ \\ dr = ds = > 1 \\ cr = cq = > 2 \\ bp = bq = > 3 \\ ap = as = > 4 \\ adding \: 1 \: 2 \: 3 \: 4 \\ (dr + cr) + (bp + ap ) = (ds + cq) + (bq + as) \\ bc \: \: \: \: \: \: \: \: \: \: \: \: + \: \: \: ad \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: \: \: \: \: dc \: \: \: \: \: \: \: + \: \: \: \: ba \\ hence \: proved$

from the figure