Question

In the given figure, `angle`ABC = `angle`ACB and `(BC)/(BE) = (BD)/(AC)`.

Show that `triangle`ABE ∼ `triangle`DBC and AE || DC.

Answer 1

Given, `angle`ABC = `angle`ACB

∴ AB = AC   ...[∵ sides opposite to equal angles of a triangle are equal]

In `triangle`ABE and `triangle`DBC,

`angle`ABE = `angle`DCB    ...[Common]

`(BC)/(BE) = (BD)/(AC)`     ...[Given]

`(BC)/(BE) = (BD)/(AB)`     ...[∵ AB = AC]

So, `triangle`ABC ∼ `triangle`DBE

Now, `(BC)/(BE) = (BD)/(AB)`

Then, by the converse of the Basic Proportionality Theorem, if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

∴ AE || AD