Question

how to solve it ????????​

Answer 1

Step-by-step explanation:

In triangle ADC

AC^2=AD^2+CD^2

but from the diagram we know that

CD=BC-BD

now sub the value of CD in the above equation

AC^2=AD^2+(BC-CD)^2

AC^2=AD^2+BC^2+BD^2-2BC.BD

but AD^2+BD^2=AB^2

therefore ,

AC^2=AB^2+BC^2-2BC.BD

HOPE IT HELPED YOU....

Answer 2

In ∆ABD,

AD² = AB²-BD² >> (i)

In ∆ACD,

AD² = AC²-DC² >> (ii)

Subtracting (i)&(ii),

0 = AB²-AC²-BD²+DC²

AC² = AB²+DC²-BD²

AC² = AB²+(DC+BD)²+2BD.DC

AC² = AB²+BC²+2BD.DC