Question

In the figure, seg AD ⊥ bC and B – D – C, then prove that:

AB2 – BD2 = AC2 – CD2
.​

Answer 1

Step-by-step explanation:

In ∆ ABD

angle ADB = 90°

using Pythagoras theorem

AB^2 = AD^2 + BD^2

AB^2 - BD^2 = AD^2

In ∆ACD

angle ADC = 90°

using Pythagoras theorem

AC^2 = AD^2 + CD^2

AC^2 - CD^2 = AD^2

As

AD^2 = AD^2

putting their values

AB^2 - BD^2 = AC^2 - CD^2

Hence Proved

LHS = RHS

I hope this will help you