{"id":11508,"date":"2022-07-12T23:10:34","date_gmt":"2022-07-12T17:40:34","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11508"},"modified":"2022-07-12T23:10:39","modified_gmt":"2022-07-12T17:40:39","slug":"in-an-equilateral-triangle-abc-d-is-a-point-on-the-side-bc-such-that-bd-1over-3-bc-prove-that-9ad2-7ab2","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-an-equilateral-triangle-abc-d-is-a-point-on-the-side-bc-such-that-bd-1over-3-bc-prove-that-9ad2-7ab2\/","title":{"rendered":"In an equilateral triangle ABC, D is a point on the side BC such that BD = \\(1\\over 3\\) BC. Prove that \\(9{AD}^2\\) = \\(7{AB}^2\\)."},"content":{"rendered":"
Let ABC be an equilateral triangle and let D be a point on BC such that<\/p>\n
BD = \\(1\\over 3\\) BC.<\/p>\n
Draw AE \\(\\perp\\) BC. Join AD.<\/p>\n
In \\(\\triangle\\) AEB and AEC, we have :<\/p>\n
AB = AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (ABC is equilateral)<\/p>\n
\\(\\angle\\) AEB = \\(\\angle\\) AEC<\/p>\n
and AE = AE<\/p>\n
\\(\\therefore\\)\u00a0 By SAS criteria of congruence, we have :<\/p>\n
\\(\\triangle\\) AEB \\(\\cong\\) \\(\\triangle\\) AEC<\/p>\n
So,\u00a0 BE = EC<\/p>\n
Now, we have :<\/p>\n
BD = \\(1\\over 3\\)BC, DC = \\(2\\over 3\\)BC and BE = EC = \\(1\\over 2\\)BC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 ……….(1)<\/p>\n
Since, \\(\\angle\\) C = 60, therefore<\/p>\n
\\(\\triangle\\) ADC is an acute triangle.<\/p>\n
\\(\\therefore\\)\u00a0 \\({AD}^2\\) = \\({AC}^2\\) + \\({DC}^2\\) – \\(2DC \\times EC\\)<\/p>\n
= \\({AC}^2\\) + \\(({2\\over 3}BC)^2\\) – \\(2\\times {2\\over 3}BC\\times {1\\over 2}BC\\)<\/p>\n
= \\({AC}^2\\) + \\(({4\\over 9}BC)^2\\) – \\(({2\\over 3}BC)^2\\)<\/p>\n
= \\({AB}^2\\) + \\(({4\\over 9}AB)^2\\) – \\(({2\\over 3}AB)^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(AB = AB = AC)<\/p>\n
= \\((9 – 4 – 6){AB}^2\\over 9\\) = \\({7\\over 9}{AB}^2\\)<\/p>\n
So,\u00a0 \\(9{AD}^2\\) = \\(7{AB}^2\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Let ABC be an equilateral triangle and let D be a point on BC such that BD = \\(1\\over 3\\) BC. Draw AE \\(\\perp\\) BC. Join AD. In \\(\\triangle\\) AEB and AEC, we have : AB = AC\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (ABC is equilateral) \\(\\angle\\) AEB = \\(\\angle\\) AEC and AE …<\/p>\n