Question

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

Answer 1

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.

Step-by-step explanation:

Referring to the figure attached below,

Considering ΔABC and ΔACD,  we have

AE is the bisector of ∠BAC

AF is the bisector of ∠CAD

We know that according to the internal bisector theorem, the angle bisector of a triangle divides the opposite sides in the ratio of sides consisting of the angles

$\frac{AC}{AB} = \frac{CE}{BE}$ …….. (i)

And

$\frac{AC}{AD} = \frac{CF}{FD}$

⇒ $\frac{AC}{AB} = \frac{CF}{FD}$ ……. [given side AB = side AD] …… (ii)

From eq. (i) & (ii), we get

$\frac{CE}{BE} = \frac{CF}{FD}$ …. (iii)

Now,  

In ΔBCD we have -  

$\frac{CE}{BE} = \frac{CF}{FD}$ ….. [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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Also View:

In a quadrilateral ABCD, AB is parallel to CD. DE and CE bisects Angle ADC and Angle BCD respectively. Prove that AB is equal to the sum of AD and BC.

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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.

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Answer 2

Answer:

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