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JOMO 5, Long 1

Triangle \(\triangle ABC\) has right angle at \(C\) such that \(AC^2+BC^2=AB^2\). Squares \(ACDE\) and \(BCFG\) are constructed on the exterior of of \(ABC\). Show that \(DF = AB\)

Note by Aditya Raut
2 years, 10 months ago

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\( \angle DCA = \angle FCB = 90^{\circ} \because ACDE \) and \( BCFG \) are both squares.

\( \angle DCA + \angle FCB + \angle ACB + \angle DCF \ = 360^{\circ} \Rightarrow \angle DCF = 90^{\circ} \)

\( AC = AE = ED = DC, \; \; BC = CF = FG = GB \)

\( \Rightarrow \triangle DCF \) is congruent to \( \triangle ABC \) as \( \angle DCF = \angle ACB, \; AC = DC, \; BC = FC \)

\( \therefore DF = AB\), as required. Danny He · 2 years, 10 months ago

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