In aIn a ABC- if D is a point on BC such that BDDC - ABAC- Prove that AD is the bisector of - A-

Construction: Produce BA to E, such that AE=AC. Join EC. Given: nbsp;AE=AC nbsp;BDDC=ABACTherefore, nbsp;BDDC=ABAE.Or nbsp;BDDC=BAAE.Therefore, nbs ...

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Standard X

Mathematics

Congruent Figures

In aIn a ABC,...

Question

In a $In a △ ABC, if D is a point on BC such that \frac{BD}{DC} = \frac{AB}{AC}, Prove that AD is the bisector of ∠ A .$

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Solution

Construction: Produce BA to E, such that AE=AC. Join EC.

$Given: AE = AC \frac{BD}{DC} = \frac{AB}{AC} Therefore, \frac{BD}{DC} = \frac{AB}{AE} . Or \frac{BD}{DC} = \frac{BA}{AE} . Therefore, according to the converse of basic proportionality theorem, AD ∥ EC . ∴ ∠ DAC = ∠ ECA [Alternate angles] . . . (1) ∴ ∠ BAD = ∠ CEA [Corresponding angles] . . . (2) G iven, AC = AE . Therefore, ∠ ECA = ∠ CEA [Angles opposite equal sides are equal] From (1) & (2), we get : ∠ DAC = ∠ BAD Hence, AD is the bisector of ∠ A .$

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Similar questions

Q. In ∆ABC (Fig. 4.59), if ∠1 = ∠2, prove that

\frac{AB}{AC} = \frac{BD}{DC}

Q. In Fig.,

D

is a point on side

B C

△ A B C

such that

\frac{B D}{C D} = \frac{A B}{A C} .

Prove that

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Q. ∆ABC is a right triangle right-angled at A and AD ⊥ BC. Then,

\frac{BD}{DC} =

(a)

{(\frac{AB}{AC})}^{2}

(b)

\frac{AB}{AC}

(c)

{(\frac{AB}{AD})}^{2}

(d)

\frac{AB}{AD}

In the given figure, D is a point on side BC of ΔABC such that. Prove that AD is the bisector of ∠BAC.

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A D^{2}

= 13

B C^{2}

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In aIn a △ABC, if D is a point on BC such that BDDC=ABAC, Prove that AD is the bisector of ∠A.

In a $In a △ ABC, if D is a point on BC such that \frac{BD}{DC} = \frac{AB}{AC}, Prove that AD is the bisector of ∠ A .$