Question

ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F, respectively. Prove that EF || BD.

In a quadrilateral ABCD, shown in the following figure, AB = AD. The bisectors of ∠BAC and ∠CAD meet the sides BC and CD at the points E and F, respectively. Prove that EF || BD.

Answer 1

ABCD is a quadrilateral, AB = AD.

AE and AF are the internal bisectors of ∠BAC and ∠DAC

.
To prove: EF || BD.

Construction: Join EF and BD

Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.

By the angle bisector theorem, we have,

∴ `"AB"/"AC" = "BE"/"EC"`   ...(1)

In ∆ ADC, AF is the internal bisector of ∠DAC.

By the angle bisector theorem, we have,

`"AD"/"AC"= "DF"/"FC"`

∴ `"AB"/"AC" = "DF"/"FC"`    ...(AB = AD given) ...(2)

From (1) and (2), we get,

`"BE"/"EC" = "DF"/"FC"`

Hence, in ∆ BCD,

BD || EF ...(by converse of BPT)

Hence proved.