Question

(f) In the given figure. 2, Take any point o in the interior of a triangle DEF Show that (i) DP + PE > DE (ii) PE+ PF > EF (iii) DP + PF > DF? ​

Answer 1

Answer:

Let O be a point in the interior of △ABC, and let OD⊥BC,OE⊥CA and OF⊥AB

(i) In right triangles △OFA,△ODB and △OEC we have

         OA2=AF2+OF2

        OB2=BD2+OD2

and, OC2=CE2+OE2

Adding all these results, we get

     OA2+OB2+OC2=AF2+BD2+CE2+OF2+OD2+OE2

⇒  AF2+BD2+CE2=OA2+OB2+OC2−OD2−OE2−OF2       [Hence proved]

(ii) In right triangles △ODB and △ODC, we have

        MARK AS BRAINLIST ANSWER