Q. 205.0( 1 Vote )

# In Fig. 10.142, if *AC* is bisector of ∠*BAD* such that *AB*=3 cm and *AC*=5 cm, then *CD*=

A. 2 cm

B. 3 cm

C. 4 cm

D. 5 cm

Answer :

In using Pythagoras theorem, we get

AB^{2} + BC^{2} = AC^{2}

9 + BC^{2} = 25

BC = 4 cm

In

∠BAC = ∠CAD (Therefore, AC is bisector of ∠A)

∠B = ∠D = 90^{o}

∠ABC + ∠BCA + ∠CAB = 180^{o}

∠CAD + ∠ADC + ∠DCA = 180^{o}

∠ABC + ∠BCA + ∠CAB = ∠CAD + ∠ADC + ∠DCA

∠BCA = ∠DCA (i)

In

∠CAB = ∠CAD (Therefore, AC is bisector of ∠A)

∠BCA = ∠DCA [From (i)

AC = AC (Common)

By ASA theorem, we have

BC = CD (By c.p.c.t)

CD = 4cm

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