In the given figure, ABC is a triangle in which ∠ABC> 90° and AD ⊥ CB produced. Prove that AC² = AB² + BC² + 2BC.BD.

Open in App

Solution

Applying Pythagoras theorem in ΔADB, we obtain

AB² = AD² + DB² … (1)

Applying Pythagoras theorem in ΔACD, we obtain

AC² = AD²+ DC²

AC² = AD² + (DB + BC)²

AC² = AD² + DB² + BC² + 2DB × BC

AC² = AB² + BC² + 2DB × BC [Using equation (1)]

Suggest Corrections

Similar questions

In the given figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC² = AB² + BC² − 2BC.BD.

In fig., $A B C$ is a triangle in which $∠ A B C > 90^{o}$ and $A D ⊥ C B$ produced. Prove that $A C^{2} = A B^{2} + B C^{2} + 2 B C . B D$ .

A B C

∠ A B C < 90^{o}

A D ⊥ B C .

A C^{2} = A B^{2} + B C^{2} - 2 B C . B D

A B C

∠ A B C < 90^{\circ}

A D ⊥ B C .

A C^{2} = A B^{2} + B C^{2} - 2 B C . D B .

A B C

∠ A B C > 90^{\circ}

A D ⊥ C B

A C^{2} = A B^{2} + B C^{2} + 2 B C . D B

Join BYJU'S Learning Program