Question

a circle touches all the four sides of equilateral ABCD prove that AB + CD is equal to BC + DA

Answer 1

Answer:

Step-by-step explanation:

0 zero

Answer 2

Proof : Let the circle touch the sides AB,BC,CD,DA of []ABCD at points P,Q,R,S respectively.
.
. . AP =AS , DS = DR ,CR = CQ ,BQ=BP (1)
and A-P-B, B-Q-C,C-R-D,A-S-D (2)
Now, AB + CD = AP + PB + CR + RD (A-P-B and
C-R-D)
= AS + BQ + CQ + DS
= AS + DS + BQ + CQ
= AD + BC (A-S-D and B-Q-C)
Thus, AB + CD= AD + BC