Question

Solve the question plzz...​

Answer 1

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✤ Required Answer:

✒ GiveN:

• AB = AD
• AE and AF are the angle bisectors of ∠BAC and ∠DAC .

✒ To Prove:

• EF || BD

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✤ How to solve?

For proving the above question, we need to know certain theorems, Based on triangles.

Thales theoram/Basic proportionality theoram:

If a line is drawn running parallel to the facing opposite side, or to one side and intersecting the other two sides, then it divides the two sides in the same ratio.

Internal bisector theoram:

The Internal angle bisector of any triangle divides the opposite facing side in the same ratio of the other two sides of the triangle.

↘️ We will use these theorems to order to prove the above$.$

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✤ Solution:

We have,

• AB = AD
• AE and AF are the bisectors of respective ∠s

In ΔBAC,

• AE is the bisector of ∠BAC

By Internal angle bisector theorem,

➙ AC/AB = CE/BE......(1)

In ΔACD,

• AF is the bisector of ∠CAD

By internal angle bisector theoram,

➙ AC/AD = CF/DF........(2)

Given ATQ,

• AB = AD

Then,

➙ Eq.(1) = Eq.(2)

➙ CE/BE = CF/DF

➙ CE/EB = CF/FD

So, E and F are the points of BC and DC which divides theses sides in equal ratio.

Now, by converse thales theoram,

➙ EF || BD

☀️ Hence, proved !!

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Answer 2

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$\large\bf \underline{GiveN:}$

• AB = AD
• E and AF are the angle bisectors of ∠BAC and ∠DAC .

To Prove:

EF || BD

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$\large\bf\underline{SoluTiOn}$

We have,

AB = AD

AE and AF are the bisectors of respective ∠s

• In ΔBAC,
• AE is the bisector of ∠BAC

By Internal angle bisector theorem,

➙ AC/AB = CE/BE......(1)

In ΔACD,

• AF is the bisector of ∠CAD

By internal angle bisector theoram,

➙ AC/AD = CF/DF........(2)

Given ATQ,

• AB = AD

Then,

$\rightarrow$Eq.(1) = Eq.(2)

$\rightarrow$ CE/BE = CF/DF

$\rightarrow$CE/EB = CF/FD

So, E and F are the points of BC and DC which divides theses sides in equal ratio.

Now, by converse thales theoram,

➙$\boxed {EF\: || \:BD}$

☄Hence, proved !!

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