In quadrilateral ABCD, side AB congruents to side AD. bisector of Angle bac cuts side BC At E and bisector of angle DAC cuts side DC at F. Prove that segEF parallel segBD.
Answers
If AB = AD and the bisector of ∠BAC and ∠DAC intersect the sides BC and DC at the points E and F respectively, then seg EF || seg BD is proved.
Step-by-step explanation:
Referring to the figure attached below,
Consider ΔABC and ΔACD,
AE is the bisector of ∠BAC
AF is the bisector of ∠CAD
We know that according to the internal bisector theorem which states that the angle bisector of a triangle divides the opposite sides in the ratio of sides consisting of the angles
AC/AB = CE/BE …….. (i)
And
AC/AD = CF/FD
⇒ AC/AB = CF/FD ……. [given side AB = side AD] …… (ii)
From eq. (i) & (ii), we get
CE/BE = CF/FD …. (iii)
Now,
In ΔBCD we have -
CE/BE = CF/FD ….. [from eq. (iii)]
We know that according to the converse of BPT theorem, if a line divides any two sides of a triangle in the same ratio, then the line should be parallel to its third side.
∴ seg EF || seg BD
Hence Proved
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