In the figure given below AB = 3 cm, AC = 5 cm and AD = 4 cm and D is the midpoint of BC. Then find the length of BD.

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Solution

Since D is the mid-point of BC,

From Apollonius theorem,

$A B^{2} + A C^{2} = 2 (B D^{2} + A D^{2}) 3^{2} + 5^{2} = 2 (B D^{2} + 4^{2}) (\frac{9 + 25}{2}) = B D^{2} + 16 \Rightarrow B D^{2} = 17 - 16 \Rightarrow B D = 1 c m$ .

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Q. In given figure, check whether AD is the bisector of

∠

A of

△

ABC in each of the following:

(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm

(ii) AB = 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2.4 cm

(iii) AB = 8 cm, AC = 24 cm, BD = 6 cm and BC = 24 cm

(iv) AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 2 cm

(v) AB = 5 cm, AC = 12 cm, BD = 2.5 cm and BC = 9 cm

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

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In the figure given below AB = 3 cm, AC = 5 cm and AD = 4 cm and D is the midpoint of BC. Then find the length of BD.

Since D is the mid-point of BC, From Apollonius theorem, AB2+AC2=2(BD2+AD2)32+52=2(BD2+42)(9+252)=BD2+16⇒ BD2=17−16⇒ BD=1 cm.

Since D is the mid-point of BC,

From Apollonius theorem,

$A B^{2} + A C^{2} = 2 (B D^{2} + A D^{2}) 3^{2} + 5^{2} = 2 (B D^{2} + 4^{2}) (\frac{9 + 25}{2}) = B D^{2} + 16 \Rightarrow B D^{2} = 17 - 16 \Rightarrow B D = 1 c m$ .