Question

P, Q and R are the mid-point of AO, BO and CO respectively as shown in figure. Prove that $\Delta ABC$ and $\Delta PQR$ are equiangular.

Answer 1

(i) In triangle OAB, P and Q are mid-points of OA and OB.
$\therefore$ From midpoint theorem PQ || BA
$\Rightarrow \angle ABO=\angle PQO$ and $\angle OAB=\angle OPQ$
(ii) In triangle OBC, Q and R are midpoints of OB and OC
$\Rightarrow QR||BC\Rightarrow \angle OBC=\angle OQR$ and $\angle OCB=\angle ORQ$
(iii) In triangle OCA, P and R are midpoints of OA and OC
$\Rightarrow PR||AC\Rightarrow \angle OCA=\angle ORP$ and $\angle OAC=\angle OPR$
Now $\angle BAC=\angle OAB+\angle OAC$
$=\angle OPQ+\angle OPR$ From (i) and (iii)
$=\angle QPR$ (iv)
$\angle ABC=\angle ABO+\angle OBC$
$=\angle PQO+\angle OQR$ From (i) and (ii)
$=\angle PQR$ (v)
Similarly, $\angle ACB=\angle PRQ$ (vi)
$\therefore \Delta ABC$ and $\Delta PQR$ are equiangular from (iv), (v) and (vi)