# In triangle ABC, AD is the bisector of the angle A and AD meets the side BC at D. If AB=5.6cm, BC=6cm and BD=3.2cm. What is the length of the side Ac ?

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Given : AD is the bisector of the angle A meeting the side BC at point D , AB=5.6cm. , Bc=6cm. and BD=3.2cm. Let us draw from point D a line (DE) parallel to the side DA meeting AC at point E.

angle BAD=angle DAE [ AD bisector of angle A ]

angle BAD=angle ADE [alternate angles equal, BAparallel to DE ]

therefore angle DAE=angle ADE and Triangle ADE is an isosceles triangle, where the side **AE=DE** , Let **AE = DE = x**

Next in triangle EDC and triangle ABC : angle BAC=angle DEC [corrosponding angles equal,BA parallel to DE] , angle BCA=angle ECD [ common angle] , angle ABD=angle EDC [ corrosponding angles equal, BA parallel to DE] . Therefore triangle ABC ~ triangle ADE [test A.A.A] . Now Using Propprtionality property of similar triangle We get :

**AB/BC = ED/AB = EC/AC** -> 2.8/6 = x/5.6 =EC/AC ------(1)

Taking x/5.6 = 2.8/6 -> x = (2.8*5.6)/6 --> x = 196/75 ---(2)

Next Taking the ratio : DC/BC = EC/AC

2.8/6 = EC/AC ---> 2.8/6 =(AC - AE)/AC [since AC= AE + EC]

2.8/6 = AC/AC - AE/AC ---> 2.8/6 = 1 - AE/AC

OR, 2.8/6 = 1 -x/AC

Or, x/AC = 1 - 2.8/6 = (6-2.8)/6 = 3.2/6

Or, X/AC = 3.2/6

OR, AC = (6*x)/3.2 = (6*196)/(3.2*75) [ substituting value of x ]

Or, Ac = (60*196)/(32*75) = 49/10 = 4.9

**Hence AC = 4.9 <--- Answer**