Im a triangle ABC ABC, AB = AC, D and E are points on the sides AB and AC respectively
In a triangle ABC, AB -
such that BD = CE. Show that:
(i) ADBC = AECB
(ii) ZDCB = ZEBC
OB-OC, where O is the point of intersection of BE and CD.
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In triangle ABC,
DE||BC
triangle ADE is similar to triangle ABC
Therefore, AD/AB = DE/BC
5/9 = DE/BC -Equation 1
In triangle DEF and BFC
angle EDF = angle BCF (alternate angles)
angle DFE = angle CFB (vertically opposite angles)
so, triangle DEF is similar to triangle BFC
Therefore, arDEF/arBFC
DE square/BC square = 5*5/9*9
=25/81
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