Like Triangles, Then take this!!

Geometry Level 3

Consider two triangles ABC and PQR such that ∠ A D B = ∠ B D C = ∠ C D A = 12 0 o \angle ADB=\angle BDC=\angle CDA=120^{o} In order to rotate the two figures we rotate PRQM by an angle θ = π 3 \theta=\frac{\pi}{3} clockwise about Q to obtain RR'QM' . Again let X and Y be points on RQ and PQ such that M lies on XY parallel to PR;

∠ R X M = 12 0 o \angle RXM=120^{o} M X = C D = w MX=CD=w ,MQM'R is concyclic R Q = u RQ=u ∠ M Q M ′ = 6 0 o \angle MQM'=60^{o} , MRR' is a right angled triangle, PYR'R is concyclic R Q = v + w + x RQ=v+w+x ∠ R X M ′ = 12 0 o \angle RXM'=120^{o} R X = B D = v RX=BD=v ,MQM'X is concyclic R Q = u + v + w RQ=u+v+w ∠ R X M ′ = 12 0 o \angle RXM'=120^{o} M ′ X = A D = u M'X=AD=u ,MQYM' is concyclic X Q = u + v XQ=u+v

Milun Moghe by Milun Moghe , 25, India

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