Question

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Answer 1

Answer: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

…….. (i)

And

⇒   ……. [given side AB = side AD] …… (ii)

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

 …. (iii)

Now,  

In ΔBCD we have -  

….. [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.

∴ segment EF // segment BD  

Hence Proved

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Answer 2

Answer:

Step-by-step explanation:

To solve this question we need to know two theorems which are-:

• Internal bisector theorem states that the angle bisector divides the opposite sides in ratio of sides consisting of the angles.

• Converse of BPT theorem states that if a line divides any two sides of a triangle in the same ratio, then the line should be parallel to its third side.

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In ΔABC AE is the bisector of ∠BAC.

AC/AB = CE/BE -- 1 ( Using Internal bisector theorem)

In ΔACD AF is the bisector of ∠CAD

AC/AD = CF/FD

AC/AB = CF/FD -- 2   {Given AD=AB}

Equating 1 and 2 we get-:  

CE/BE = CF/FD

In ΔBCD we have -  

CE/BE = CF/FD

Therefore, using converse of BPT Theorem.

= EF || BD  [Hence Proved]