Question

. In the adjacent figure in which AD BC  and AB = p, AC = q, BD = r, DC = s then

show that p²-q²=r²-s²​

Answer 1

Answer:

In ΔABD,

(AB)²=(AD)²+(BD)²

(AD)²=(AB)²-(BD)²

(AD)²=p²-r²--eq1

In ΔACD,

(AC)²=(AD)²+(CD)²

(AD)²=(AC)²-(CD)²

(AD)²=q²-s²--eq2

From eq1 and eq2, we get

p²-r²=q²-s²

p²-q²=r²-s²

Hence proved