{"id":11358,"date":"2022-07-07T23:08:04","date_gmt":"2022-07-07T17:38:04","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11358"},"modified":"2022-07-07T23:08:37","modified_gmt":"2022-07-07T17:38:37","slug":"e-is-point-on-the-side-ad-produced-of-a-parallelogram-abcd-and-be-intersects-cd-at-e-show-that-triangle-abe-triangle-cfb","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/e-is-point-on-the-side-ad-produced-of-a-parallelogram-abcd-and-be-intersects-cd-at-e-show-that-triangle-abe-triangle-cfb\/","title":{"rendered":"E is point on the side AD produced of a parallelogram ABCD and BE intersects CD at E. Show that \\(\\triangle\\) ABE ~ \\(\\triangle\\) CFB"},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>In triangles ABE and CFB, we have,<img decoding=\"async\" loading=\"lazy\" class=\" wp-image-11359 alignright\" src=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-8.jpeg?resize=213%2C150&#038;ssl=1\" alt=\"\" width=\"213\" height=\"150\" srcset=\"https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-8.jpeg?resize=300%2C211&amp;ssl=1 300w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-8.jpeg?resize=1024%2C722&amp;ssl=1 1024w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-8.jpeg?resize=768%2C541&amp;ssl=1 768w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-8.jpeg?resize=1536%2C1082&amp;ssl=1 1536w, https:\/\/i0.wp.com\/mathemerize.com\/wp-content\/uploads\/2022\/07\/similartriangles6-3-8.jpeg?w=1700&amp;ssl=1 1700w\" sizes=\"(max-width: 213px) 100vw, 213px\" data-recalc-dims=\"1\" \/><\/p>\n<p>\\(\\angle\\) AEB = \\(\\angle\\) CBF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(alternate \\(\\angle\\)s are equal)<\/p>\n<p>\\(\\angle\\) A = \\(\\angle\\) C\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (opposite angles of parallelogram are equal)<\/p>\n<p>\\(\\therefore\\) By AA similarity,<\/p>\n<p><strong>\\(\\triangle\\) ABE ~ \\(\\triangle\\) CFB<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : In triangles ABE and CFB, we have, \\(\\angle\\) AEB = \\(\\angle\\) CBF\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(alternate \\(\\angle\\)s are equal) \\(\\angle\\) A = \\(\\angle\\) C\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (opposite angles of parallelogram are equal) \\(\\therefore\\) By AA similarity, \\(\\triangle\\) ABE ~ \\(\\triangle\\) CFB<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,912],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>E is point on the side AD produced of a parallelogram ABCD and BE intersects CD at E. 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