{"id":11294,"date":"2022-07-01T01:27:29","date_gmt":"2022-06-30T19:57:29","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11294"},"modified":"2022-07-01T01:27:32","modified_gmt":"2022-06-30T19:57:32","slug":"prove-that-the-line-drawn-from-the-mid-point-of-one-side-of-triangle-parallel-to-another-side-bisects-the-third-side","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/prove-that-the-line-drawn-from-the-mid-point-of-one-side-of-triangle-parallel-to-another-side-bisects-the-third-side\/","title":{"rendered":"Prove that the line drawn from the mid-point of one side of triangle parallel to another side bisects the third side."},"content":{"rendered":"
Given<\/strong> : A triangle ABC in which D is the mid point of side AB and the line DE is drawn parallel to BC, meeting AC in E.<\/p>\n To Prove<\/strong> :\u00a0 AE = EC<\/p>\n Proof<\/strong> : In triangle ABC, DE || BC<\/p>\n By basic proportionality theorem,<\/p>\n \\(AD\\over DB\\) = \\(AE\\over EC\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ………..(1)<\/p>\n But given that D is the mid point of AB,<\/p>\n AB = DB\u00a0 \u00a0 \u00a0\\(\\implies\\)\u00a0 \u00a0\\(AD\\over DB\\) = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0…….(2)<\/p>\n From (1) and (2), we obtain that<\/p>\n \\(AE\\over EC\\) = 1\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \\(\\implies\\)\u00a0 \u00a0AE = EC.<\/p>\n Hence, E bisects AC.<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : A triangle ABC in which D is the mid point of side AB and the line DE is drawn parallel to BC, meeting AC in E. To Prove :\u00a0 AE = EC Proof : In triangle ABC, DE || BC By basic proportionality theorem, \\(AD\\over DB\\) = \\(AE\\over EC\\)\u00a0 \u00a0 \u00a0 …<\/p>\n