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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. · 2 years, 10 months ago

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