{"id":11407,"date":"2022-07-08T22:27:12","date_gmt":"2022-07-08T16:57:12","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11407"},"modified":"2022-07-08T22:27:15","modified_gmt":"2022-07-08T16:57:15","slug":"in-the-figure-abc-and-dbc-are-two-triangles-on-the-same-base-bc-if-ad-intersects-bc-at-o-show-that-areatriangle-abcover-areatriangle-dbc-aoover-do","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-the-figure-abc-and-dbc-are-two-triangles-on-the-same-base-bc-if-ad-intersects-bc-at-o-show-that-areatriangle-abcover-areatriangle-dbc-aoover-do\/","title":{"rendered":"In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\)."},"content":{"rendered":"
Given<\/strong> : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC.<\/p>\n To Prove<\/strong> : \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\)<\/p>\n Construction<\/strong> : Draw AE \\(\\perp\\) BC and DF \\(\\perp\\) BC.<\/p>\n Proof<\/strong> : In triangles AOE and DOF, we have :<\/p>\n \\(\\angle\\) AEO = \\(\\angle\\) DFO = 90<\/p>\n \\(\\angle\\) AOE = \\(\\angle\\) DOF\u00a0 \u00a0 \u00a0 \u00a0(vertically opposite angles)<\/p>\n By AA similarity, we have :<\/p>\n \\(\\triangle\\) AOE ~ \\(\\triangle\\) DOF<\/p>\n So, \\(AE\\over DF\\) = \\(AO\\over OD\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……..(1)<\/p>\n Now, \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\({1\\over 2} \\times BC \\times AE\\over {1\\over 2}\\times BC \\times DF\\) = \\(AE\\over DF\\) = \\(AO\\over OD\\)<\/p>\n \\(\\therefore\\)\u00a0 \u00a0 \u00a0\\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over OD\\)<\/strong>.<\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Given : Two triangles ABC and DBC which stand on the same base BC but on the opposite sides of BC. To Prove : \\(area(\\triangle ABC)\\over area(\\triangle DBC)\\) = \\(AO\\over DO\\) Construction : Draw AE \\(\\perp\\) BC and DF \\(\\perp\\) BC. Proof : In triangles AOE and DOF, we have : \\(\\angle\\) AEO …<\/p>\n