plz tell the solution for this question its very important....
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we will use the property of congruent triangles
given AB=DE
angle BCA= angle ECD(vertically opposite)
angle ABC=angle CDE(right angled)
thus by ASA the triangles ABC and CDE are congruent
thus BC=CD
given AB=DE
angle BCA= angle ECD(vertically opposite)
angle ABC=angle CDE(right angled)
thus by ASA the triangles ABC and CDE are congruent
thus BC=CD
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as...
AB=DE(given)
angleB=angleD(each 90°)
AC=CE( each hypotenuse)
therefore ΔABC congruent to ΔCDE(by RHS)
so,CD=BC(C.P.C.T)corresponding part of congruent Δ....
☜☆☞☜☆☞☜☆☞╭∩╮（︶︿︶）╭∩╮
AB=DE(given)
angleB=angleD(each 90°)
AC=CE( each hypotenuse)
therefore ΔABC congruent to ΔCDE(by RHS)
so,CD=BC(C.P.C.T)corresponding part of congruent Δ....
☜☆☞☜☆☞☜☆☞╭∩╮（︶︿︶）╭∩╮
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