In the given figure, if $A D ⊥ B C$ and BD = DC, then AD bisects $∠ B A C$ .

True

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False

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Solution

The correct option is A
True

In $△ B A D ≅ △ C A D$

(i) AD = AD ......(common)

(ii) $∠ A D B = ∠ A D C = 90^{\circ}$

(iii) BD = CD ......(given)

(iv) $△ B A D ≅ △ C A D$ ........(SAS postulate)

(v) $∠ B A D = ∠ C A D$

Then AD bisects $∠ B A C$ ........(from (v))

Hence the given statement is true.

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Similar questions

∠ B A C = 90^{0}

A D ⊥ B C

{A D}^{2} = B D \times D C

In the given figure, if $A D ⊥ B C$ and BD = DC, then which of the following are true ?

Δ A B C

∠ B A C

A D = D C .

∠ A D B = 100^{0},

∠ A B D .

In figure, $∠ B A C = 90^{\circ}$ and segment $A D ⊥ B C$ . Prove that ${A D}^{2} = B D \times D C$ . [3 MARKS]

∠ B A C

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