Math, asked by srijajana674, 1 year ago

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.​

Answers

Answered by Anonymous
1

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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