# In the given figure, ΔABC is an obtuse triangle, obtuse angled at B. If AD⊥CB , then prove that AC^{2} = AB^{2} + BC^{2} + 2BC**.**BD

**Solution:**

From the right angled triangle ADC,

Using pythagoras theorem, AD^{2} + DC^{2} = AC^{2} --- (1)

From the figure,

DC = DB +BC --- (2)

Substitute DC from (2) in equation (1)

AD^{2} + (BD + BC)^{2} = AC^{2}

AC^{2} = AD^{2} + BD^{2} + BC^{2} + 2BD.BC --- (3)

Also from right angled triangle ADB,

AD^{2} + BD^{2} = AB^{2} (From pythagoras theorem) --- (4)

Substitute (4) in equation (3)

AC^{2} = AB^{2} + BC^{2} + 2BD.BC

## In the given figure, ΔABC is an obtuse triangle, obtuse angled at B. If AD⊥CB , then prove that AC^{2} = AB^{2} + BC^{2} + 2BC**.**BD ?

**Summary:**

Given ΔABC is an obtuse triangle, obtuse angled at B and AD⊥CB, then AC^{2} = AB^{2} + BC^{2} + 2BC**.**BD.

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