Question

SOT
In ABCD, side AB side AD. Bisector
of ZBAC cuts side BC at E and bisector
of ZDAC cuts side DC at F. Prove that seg EF
|| seg BD.​

Answer 1

Proved that EF║BD    , if AB = AB &  The bisector of  ∠BAC AND ∠DAC intersect the sides BC and DC at the points E and F

Step-by-step explanation:

in Δ ABC

AC/AB = CE / BE  (  Using internal bisector Theorem)

Similarly

in Δ ACD

AC/AD =  CF/FD

=> AC/AB =  CF/FD

as AB = AD

=> CE / BE = CF/FD

now in Δ BCD

CE / BE = CF/FD

=> EF║BD  ( Converse of BPT Theorem)

QED

Proved

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