In Figure, D is a point on side BC of ∆ ABC such that \( \dfrac{BD}{CD}=\dfrac{AB}{AC}\). Prove that AD is the bisector of ∠ BAC.

In the given figure, D is a point on side BC of ∆ABC, such that - CBSE  Class 10 Maths - Learn CBSE Forum

 

Asked by Pragya Singh | 1 year ago |  207

1 Answer

Solution :-

In the given let us extend BA to P such that;

AP = AC.

Now join PC.

Given, \( \dfrac{BD}{CD}=\dfrac{AB}{AC}\)

\( \dfrac{BD}{CD}=\dfrac{AP}{AC}\)

By using the converse of basic proportionality theorem, we get,

AD || PC

∠BAD = ∠APC (Corresponding angles) ……….. (i)

And, ∠DAC = ∠ACP (Alternate interior angles) …… (ii)

By the new figure, we have;

AP = AC

∠APC = ∠ACP ………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

∠BAD = ∠APC

Therefore, AD is the bisector of the angle BAC.

Hence, proved.

Answered by Abhisek | 1 year ago

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