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State the converse of Pythagoras theorem and prove it.

Last updated date: 13th Jun 2024
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Hint: We will first write the statement of the converse of Pythagoras theorem then to prove it we will first consider two triangles and prove them congruent by SSS congruency. Then we will prove that one of the angles of a triangle is a right angle.

Complete step by step answer:
We know that the converse of Pythagoras theorem is stated as:
In a triangle, if the square of one longest side is equal to the sum of squares of the other two sides then the angle opposite the first side is a right angle.
Now, to prove this statement let us consider two triangles $ \Delta ABC\And \Delta PQR $ from which $ \Delta PQR $ is right-angled at $ Q $ .

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In $ \Delta ABC $ we have $ AC $ as the longest side and by Pythagoras theorem we get
 $ \Rightarrow A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}...........(i) $
We have to prove that $ \Delta ABC $ is right triangle.
Now, we construct another triangle $ \Delta PQR $ which is a right triangle $ \angle Q=90{}^\circ $ and AB=PQ and BC=QR
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So, by Pythagoras theorem we have $ P{{R}^{2}}=Q{{R}^{2}}+P{{Q}^{2}} $
Now, by construction we have $ AB=PQ\And BC=QR $
So, substituting the values we get
 $ \Rightarrow P{{R}^{2}}=A{{B}^{2}}+B{{C}^{2}}...........(ii) $
Now, from equation (i) and (ii) we get
 $ \begin{align}
  & \Rightarrow A{{C}^{2}}=P{{R}^{2}} \\
 & \Rightarrow AC=PR \\
\end{align} $
Now, in $ \Delta ABC\And \Delta PQR $ we have
 $ \begin{align}
  & \Rightarrow AB=PQ \\
 & \Rightarrow BC=QR \\
\end{align} $ (by construction)
 $ \Rightarrow AC=PR $ (Proved above)
So we get $ \Delta ABC\cong \Delta PQR $ (SSS congruency)
So we get
 $ \angle Q=\angle B $ (corresponding angles of congruent triangles)
And we have $ \angle Q=90{}^\circ $
So, $ \angle B=90{}^\circ $
Hence proved that $ \Delta ABC $ is right triangle.

The fact is that to write the converse of a theorem we need to interchange the hypothesis and conclusion of a statement. The key point to prove triangles congruent we need to construct the two sides of the triangles equal.