3 Tutor System
Starting just at 265/hour

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

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

We can write it as;
$$AD^2 = AB^2 - DB^2$$……………….. (i)

By applying Pythagoras Theorem in $$\triangle$$ ADC, we get,
$$AD^2 + DC^2 = AC^2$$

From equation (i),
$$AB^2 - BD^2 + DC^2 = AC^2$$
$$AB^2 - BD^2 + (BC - BD)^2 = AC^2$$
$$AC^2 = AB^2 - BD^2 + BC^2 + BD^2 - 2BC × BD$$
$$AC^2= AB^2 + BC^2 - 2BC × BD$$
Hence, proved.