3 Tutor System
Starting just at 265/hour

# In Figure, ABC is a triangle in which $$\angle$$ ABC > 90° and AD $$\perp$$ CB produced. Prove that $$AC^2= AB^2+ BC^2+ 2 BC.BD.$$

By applying Pythagoras Theorem in $$\triangle$$ ADB, we get,
$$AB^2 = AD^2 + DB^2$$ ……………………… (i)

Again, by applying Pythagoras Theorem in $$\triangle$$ ACD, we get,
$$AC^2 = AD^2 + DC^2$$
$$AC^2 = AD^2 + (DB + BC)^2$$
$$AC^2 = AD^2 + DB^2 + BC^2 + 2DB × BC$$

From equation (i), we can write,
$$AC^2 = AB^2 + BC^2 + 2DB × BC$$
Hence, proved.