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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)
1378622_cee6ffc696114f30a9f089fd8694ef14.png


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.

1303696_1378622_ans_28358894258f4eadbff364282848643f.PNG

Mathematics

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