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Offline Centres

State True or False.

In the given figure, D is a point on side BC of ∆ ABC, such that

$B D C D = A B A C$ .Then AD is not the bisector of ∠BAC.

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True

False

answer is B.

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Detailed Solution

Given that, D is a point on side BC of ∆ ABC, such that

B D C D = A B A C

We extend AB such that AE = AC. Join E and C.
Question Image

Given that

B D C D = A B A C

,
Also, AE = AC.

⇒ B D C D = A B A E

…..[1][By construction]
Converse of BPT is, if a line divides the two sides of a triangle in the same ratio, then the line is parallel to the third side of the triangle.
So by converse of BPT, in

Δ ABC

, AD||CE.

∠ B A D = ∠ B E C

…..[2][Corresponding Angles]

∠ D A C = ∠ A C E

…..[3][Alternate interior angles]

Δ AEC

is an isosceles triangle by construction,

∠ A C E = ∠ A E C

…..[4][Angles opp. to equal sides]
From [2], [3] and [4], we get

∠ B A D = ∠ D A C

.
Hence, AD bisects

∠ B A C

.
The given statement is false.
In the given figure, if D is a point on side BC of ∆ ABC, such that

B D C D = A B A C

, then AD is the bisector of ∠BAC.
Therefore, the correct option is 2.

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