Q8 of 47 Page 1

The in circle of an isosceles triangles ABC, in which AB = AC, touches the sides BC, CA and AB at D, E and F respectively. Prove that BD = DC.


Given, AB = AC


⸫ AF = AE [Tangents drawn from an external point to the same circle are equal]


⸫ BD = BF [Tangents drawn from an external point to the same circle are equal]


⸫ DC = CE [Tangents drawn from an external point to the same circle are equal]


From the above 3 eqs, we can conclude


AF + BF + DC = AE + CE + BD


⇒ AB + DC = AC + BD


[⸪ we know that, AB = AC]


⇒ AC + DC = AC + BD


⸫ BD = DC


Hence Proved.


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