Plz solve this sum......
If you know then solve.....
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le behena :-)
In right-angle triangles BCE and CBF, we have,
BC = BC (common hypotenuse);
BE = CF (given).
Hence BCF and CBF are congruent, by RHS theorem. Comparing the triangles, we get ∠B=∠C.
This implies that
AC = AB (sides opposite to equal angles).
Similarly,
AD=BE⇒∠B=∠A⇒AC=BC
Together, we get AB=BC=AC or △ABC is equilateral triangle.
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