Let A,B,C be angles of triangle ABC with vertex $A \equiv (4, - 1)$ and $x - 1 = 0$ and $x - y - 1 = 0$ are internal angle bisectors through B and C respectively. Let D,E,F are respectively the points of contact of sides BC, CA and AB with incircle of $Δ A B C$ . If $D^{'}, E^{'}, F^{'}$ are images of D, E and F through internal angle bisectors of angles A,B,C respectively then which of the following option(s) is/are INCORRECT?

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Area of $Δ A B C$ is 23 sq. units

Equation of incircle of $Δ A B C$ is ${(x - 1)}^{2} + y^{2} = 5$

Circumcircle of $Δ D^{'} E^{'} F^{'}$ is ${(x - 1)}^{2} + y^{2} = 5$

Equation of incircle of $Δ A B C$ is $x^{2} + y^{2} = 5$

answer is C, D.

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Detailed Solution

Mirror images of A(4,-1) about the two bisectors x-y-1=0 and x-1=0 are respectively (0,3) and (-2,-1) which lie in BC.
Equation of BC is 2x-y+3=0 Inradius $r = \frac{5}{\sqrt{5}} = \sqrt{5}$
Again image of a point on circumcircle with respect to diameter lies on circle again i.e. circle is ${(x - 1)}^{2} + {(y - 0)}^{2} = {(\sqrt{5})}^{2}$