{"id":11479,"date":"2022-07-12T12:42:46","date_gmt":"2022-07-12T07:12:46","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11479"},"modified":"2022-07-12T12:53:02","modified_gmt":"2022-07-12T07:23:02","slug":"in-the-figure-o-is-a-point-in-the-interior-of-a-triangle-abc-od-perp-bc-oe-perp-ac-and-of-perp-ab","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-o-is-a-point-in-the-interior-of-a-triangle-abc-od-perp-bc-oe-perp-ac-and-of-perp-ab\/","title":{"rendered":"In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB."},"content":{"rendered":"
In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB. Show that<\/p>\n
(i)<\/strong>\u00a0 \\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) – \\({OD}^2\\) – \\({OE}^2\\) – \\({OF}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\)<\/p>\n (ii)<\/strong>\u00a0 \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) = \\({AE}^2\\) + \\({CD}^2\\) + \\({BF}^2\\)<\/p>\n Join AO, BO and CO as shown in fig.<\/p>\n (i)<\/strong>\u00a0 In right \\(\\triangle\\)s OFA, ODB and OEC, we have<\/p>\n \\({OA}^2\\) = \\({AF}^2\\) + \\({OF}^2\\)<\/p>\n \\({OB}^2\\) = \\({BD}^2\\) + \\({OD}^2\\)<\/p>\n and \\({OC}^2\\) = \\({CE}^2\\) + \\({OE}^2\\)<\/p>\n Adding all these, we get<\/p>\n \\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) + \\({OD}^2\\) + \\({OE}^2\\) + \\({OF}^2\\)<\/p>\n or\u00a0 \u00a0\\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) – \\({OD}^2\\) – \\({OE}^2\\) – \\({OF}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\)<\/strong><\/p>\n (ii)\u00a0 In right \\(\\triangle\\)s ODB and ODC, we have :<\/p>\n \\({OB}^2\\) = \\({BD}^2\\) + \\({OD}^2\\)\u00a0 and\u00a0 \\({OC}^2\\) = \\({OD}^2\\) + \\({CD}^2\\)<\/p>\n or\u00a0 \\({OB}^2\\) – \\({OC}^2\\) = \\({BD}^2\\) – \\({CD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……….(1)<\/p>\n Similarly, we have :<\/p>\n \\({OC}^2\\) – \\({OA}^2\\) = \\({CE}^2\\) – \\({AE}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0………..(2)<\/p>\n \\({OA}^2\\) – \\({OB}^2\\) = \\({AF}^2\\) + \\({BF}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(3)<\/p>\n Adding equations (1), (2) and (3), we get<\/p>\n \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) = \\({AE}^2\\) + \\({CD}^2\\) + \\({BF}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Question : In the figure, O is a point in the interior of a triangle ABC, OD \\(\\perp\\) BC, OE \\(\\perp\\) AC and OF \\(\\perp\\) AB. Show that (i)\u00a0 \\({OA}^2\\) + \\({OB}^2\\) + \\({OC}^2\\) – \\({OD}^2\\) – \\({OE}^2\\) – \\({OF}^2\\) = \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) (ii)\u00a0 \\({AF}^2\\) + \\({BD}^2\\) + \\({CE}^2\\) = \\({AE}^2\\) + …<\/p>\nSolution :<\/h2>\n