Question

In the given fig, D and E are points on sides AB and CA of ∆ABC such that ∠ B = ∠ AED. Show that ∆ABC ~ ∆AED.

Answer 1

Two Triangles are said to be similar if their i)corresponding angles are equal and ii)corresponding sides are proportional.(the ratio between the lengths of corresponding sides are equal)

•Similarity of triangles should be expressed symbolically using correct correspondence of their vertices

AA similarity: if two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar.

SOLUTION:

GIVEN:
∠B =  ∠AED

In the following figure we have ∆ABC & ∆AED in which,
∠ABC =  ∠AED (GIVEN)
∠BAC =  ∠EAD  (COMMON)
∆ABC ~  ∆AED

[ By AA criterion of Similarity ]

Hence, ∆ABC ~  ∆AED

HOPE THIS WILL HELP YOU.

Answer 2

angle B = angle AED  ( given)

angle DAE = angle DAE    ( common)

angle ADE = angle ACB  ( remaining angles)

hence, proved triangle ABC~AED

i hope this will help you


-by ABHAY