Question

In the adjoining figure
seg AD I seg BC, B-D-C.
Show that
AB2 + CD2 = AC2 + BD2.
In ∆ ABD, AB2 = AD2 + BD2 ..... : Pythagoras Theorem
.. AD2 = AB2 - BD2 .... (1)
In ∆ ACD, = AD2 + DC2 ....
. = AC2- .... (II)
· AB2 - = - DC.... from (1) and (II)
. AB2 + CD2 = AC2 + BD2.​

Answer 1

Step-by-step explanation:

1. AC^2 Pythagorean theorem

2. Ad^2= Ac^2- DC^2

3.AB^2-BD^2=AC^2-DC^2

Answer 2

Answer:

in Above Picture All Answers