Question

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

Answer 1

To prove : BD= DC
Construction : Join OB and OC
Proof:
In triangle ABC AB = AC
= Angle B = Angle C ..... (i)
Now O is the incenter
Hence, it lies on bisectors of angleB and angleC
AngleOBC = Half angleB .... (ii)
And angle OCB = Half angleC .... (iii)

From (i), (ii) and (iii), we get
Half angleB = Half angleC
i.e. AngleOBC= AngleOCB ..... (iv)
Now, in triangle OBD and OCD,
AngleOBD = AngleOCD ... [from (iv)]
Angle ODB = ODC ..... [Each 90degree]
And OD= OD.... [ Common side ]
Triangle OBD =~ Triangle OCD.. [SAA rule]
BD = DC ..... [CPCT]
Hence, proved

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