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Questions
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.
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Solution
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.
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