Answer:
Coordinates of vertices are \[A(-2,0),B(2,0),C(0,2)\] \[P(-4,0),Q(4,0),R(0,4)\] \[AB=\sqrt{{{(2+2)}^{2}}+{{(0-0)}^{2}}}=4\,\,units\] \[BC=\sqrt{{{(0-2)}^{2}}+{{(2-0)}^{2}}}\] \[=\sqrt{4+4}=2\sqrt{2}\,units\] \[CA=\sqrt{{{(-2-0)}^{2}}+{{(0-2)}^{2}}}\] \[=\sqrt{8}=2\sqrt{2}\,\,units\] \[PR=\sqrt{{{(0+4)}^{2}}+{{(4-0)}^{2}}}\] \[=\sqrt{{{4}^{2}}+{{(4)}^{2}}}=4\sqrt{2}\,\,units\] \[QR=\sqrt{{{(0-4)}^{2}}+{{(4-0)}^{2}}}\] \[=\sqrt{{{4}^{2}}+{{(4)}^{2}}}=4\sqrt{2}\,\,units\] \[PQ=\sqrt{{{(4+4)}^{2}}+{{(0-0)}^{2}}}\] \[=\sqrt{{{(8)}^{2}}}=8\,\,units\] We see that sides of \[\Delta \text{ }PQR\] are twice the sides of \[\Delta \text{ ABC}\]. Hence both triangles are similar. Hence Proved.
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