A(-4, 2), B(0, 2) and C(-2, -4) are vertices of a triangle ABC. P, Q and R are mid-points of sides BC, CA and AB respectively. Show that the centroid of $Δ$ PQR is the same as the centroid of $Δ$ ABC.

Q. The sides AB BC and CA of triangle ABC touch a circle with center O and radius r at P, Q and R respectively. Prove that area of triangle ABC =

\frac{r}{2} \times p e r i m e t e r o f t r i a n g l e A B C

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Let ABC be a triangle, from vertices A,B and C altitudes AD,BE and CF are drawn to opposite sides BC,CA and AB respectively which meet at a point O. Now triangle DEF is completed, also length OA,OB and OC respectively are p,q and r.Radius of the circumcircle of ΔDEF is

The correct option is A R/2Letcircumradiusbe′R′2R′=EFsin∠E−D=acosAsin(π−2A)=a2sinA=RR′=R2Hence,optionAisthecorrectanswer.

The correct option is A $R / 2$
$\begin{matrix} L e t c i r c u m r a d i u s b e^{'} R^{'} 2 R^{'} = \frac{E F}{sin ∠ E - D} = \frac{a cos A}{sin (π - 2 A)} = \frac{a}{2 sin A} = R R^{'} = \frac{R}{2} H e n c e, o p t i o n A i s t h e c o r r e c t a n s w e r . \end{matrix}$