In `DeltaABC`, ray `BD` bisects `/_ABC`. `A-D-C`, side `DE||` side `BC`, `A-E-B`. Prove that, `(AB)/(BC)=(AE)/(EB)`. Complete the activity by filling the boxes. In `DeltaABC`, ray `BD` is the bisector of `/_ABC` `:.(AB)/(BC)=square`.......`(I)` (By angle bisector theorem) In `DeltaABC`, seg `DE||` side `BC` `:.(AE)/(EB)=(AD)/(DC)`........`(II)` `square` `:.(AB)/(square)=(square)/(EB)`.......[From `(I)` and `(II)`]

In `DeltaABC`, ray `BD` bisects `/_ABC`. `A-D-C`, side `DE||` side `BC

1.	In `DeltaABC`, ray `BD` bisects `/_ABC`. `A-D-C`, side `DE\|\|` side `BC`, `A-E-B`. Prove that, `(AB)/(BC)=(AE)/(EB)`. Complete the activity by filling the boxes. In `DeltaABC`, ray `BD` is the bisector of `/_ABC` `:.(AB)/(BC)=square`.......`(I)` (By angle bisector theorem) In `DeltaABC`, seg `DE\|\|` side `BC` `:.(AE)/(EB)=(AD)/(DC)`........`(II)` `square` `:.(AB)/(square)=(square)/(EB)`.......[From `(I)` and `(II)`]
Answer» In `DeltaABC`, ray `BD` is the bisector of `/_ABC` `:.(AB)/(BC)=(AD)/(DC)`.......`(I)` (By angle bisector theorem) In `DeltaABC`, seg `DE\|\|` side `BC` `:.(AE)/(EB)=(AD)/(DC)`........`(II)` By basic proportionality theorem `:.(AB)/(BC)=(AE)/(EB)`.......[From `(I)` and `(II)`]

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