Answer :

We will have to make the following construction in the given figure:


Produce CB to D in such a way that BD=BC and join AD.


Now, in ∆ABC and ∆ABD,


BC=BD (constructed)


AB=AB (common)


ABC=ABD (each 90°)


by S.A.S.


∆ABC ∆ABD


CAB=DAB and AC=AD (by c.p.c.t.)


∴∠CAD=CAB+BAD


=x°+x°


=2x°


But, AC=AD


ACD=ADB=2x°


∆ACD is equilateral triangle.


AC=CD


AC=2BC


Hence, proved


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