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**Best Answer**

Let *ABC* represent a right triangle, with the right angle located at *C*, as shown on the figure. We draw the altitude from point *C*, and call *H* its intersection with the side *AB*. The new triangle *ACH* is similar to our triangle *ABC*, because they both have a right angle (by definition of the altitude), and they share the angle at *A*, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to *ABC*. The similarities lead to the two ratios:

These can be written as

Summing these two equalities, we obtain

In other words, the Pythagorean theorem:

- PROVED...

In the given Figure, ABC is a triangle in which ABC > 90° and AD ⊥ CB

produced. Prove that

AC^{2} = AB^{2} + BC2 + 2 BC . BD

Given : ABC is a triangle in which ABC > 90° and AD ⊥ CB produced.

To Prove: AC^{2} = AB^{2} + BC^{2} + 2 BC . BD

Proof:

AD ⊥ CB

⇒∆ADB is a right triangle , right angled at D

From Pythagoras theorem, we have

AC^{2}= AD^{2}+DB^{2}

Also, AD ⊥ CB

⇒∆ADC is a right triangle , right angled at D

From Pythagoras theorem, we have

AC^{2} = AD^{2} + DC^{2}

⇒AC^{2} = AD^{2}+(DB+BC)^{ 2}

⇒AC^{2}= AD^{2}+DB^{2} +BC^{2} +2 BC.BD

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

⇒AC^{2}= AB^{2}+BC^{2}+2 BC.BD

Thus, the theorum is proved.