In $△ A B C, A B = 3 c m, A C = 4 c m$ and AD is the bisector of $∠$ A. Is $B D = D C ?$

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Solution

Using angle bisector theorem, $\frac{B D}{D C} = \frac{A B}{A C} = \frac{3}{4}$ [1 Mark]

So, $B D : D C = 3 : 4$

Since, $B D : D C \neq 1 : 1$

Hence, $B D \neq D C$ . [1 Mark]

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Q. In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC

(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 8 cm

In a $Δ A B C$ , AD is the bisector of $∠ A$ .

(i) If AB = 6.4 cm, AC = 8 cm and BD = 5.6 cm, find DC.

(ii) If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

(iii) If AB = 5.6 cm, BD = 3.2 cm and BC = 6 cm, find AC.

(iv) If AB = 5.6 cm, AC = 4 cm and DC = 3 cm, find BC.

In a $Δ A B C$ , AD is the bisector of $∠ A$ , meeting side BC at D.
(i) If BD= 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC.
(ii) If BD= 2 cm, AB = 5 cm and DC = 3 cm, find AC.
(iii) If AB= 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.
(iv) If AB =10 cm, AC =14 cm and BC = 6 cm, find BD and DC.
(v) If AC = 4.2 cm, DC = 6 cm and BC = 10 cm, find AB.
(vi) If AD= 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
(vii) If AB= 3.6 cm, BC e 6 cm and BD = 32 cm, find AC.
(viii) If AB =10 cm, AC = 6 cm and BC = 12 cm, find BD and DC

In $△ A B C, A B = 3 c m, A C = 4 c m$ and AD is the bisector of $∠$ A. Then, $B D = D C .$

In $△$ ABC, AB = 3 and, AC = 4 cm and AD is the bisector of $∠$ A. Then, BD : DC is _____.

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In △ABC, AB=3 cm, AC=4 cm and AD is the bisector of ∠A. Is BD=DC?

​​ Using angle bisector theorem, BDDC=ABAC=34 [1 Mark] So, BD:DC=3:4 Since, BD:DC≠1:1 Hence, BD≠DC. [1 Mark]

In $△ A B C, A B = 3 c m, A C = 4 c m$ and AD is the bisector of $∠$ A. Is $B D = D C ?$

Using angle bisector theorem, $\frac{B D}{D C} = \frac{A B}{A C} = \frac{3}{4}$ [1 Mark]

So, $B D : D C = 3 : 4$

Since, $B D : D C \neq 1 : 1$

Hence, $B D \neq D C$ . [1 Mark]