Question

# In the given figure ABC is a right triangle and right angled at B such that $$\angle BCA=2\angle BAC$$.Show that hypotenuse AC = 2BC.(Hint : Produce CB to a point D that BC = BD)

Solution

## In $$\triangle{ABD}$$ and $$\triangle{ABC}$$ we have  $$BD = BC$$$$AB = AB$$ [Common]   $$\angle{ABD}=\angle{ABC}={90}^{\circ}$$$$\therefore$$ By SAS criterion of congruence we get   $$\triangle{ABD}\cong \triangle{ABC}$$  $$\Rightarrow AD = AC$$ and $$\angle{DAB}=\angle{CAB}$$ [By CPCT]  $$\Rightarrow AD = AC$$ and $$\angle{DAB}=x$$ [$$\therefore\angle{CAB}=x$$]   Now, $$\angle{DAC}=\angle{DAB}+\angle{CAB}= x + x = 2x$$$$\therefore \angle{DAC}=\angle{ACD}$$$$\Rightarrow DC = AD$$ [Side Opposite to equal angles]$$\Rightarrow 2BC = AD$$ since $$DC = 2BC$$$$\Rightarrow 2BC = AC$$ Since $$AD = AC$$Hence proved.Mathematics

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