{"id":15595,"date":"2024-02-29T06:00:58","date_gmt":"2024-02-29T00:30:58","guid":{"rendered":"https:\/\/www.aplustopper.com\/?p=15595"},"modified":"2024-02-29T15:19:31","modified_gmt":"2024-02-29T09:49:31","slug":"selina-icse-solutions-class-9-maths-mid-point-converse-including-intercept-theorem","status":"publish","type":"post","link":"https:\/\/www.aplustopper.com\/selina-icse-solutions-class-9-maths-mid-point-converse-including-intercept-theorem\/","title":{"rendered":"Selina Concise Mathematics Class 9 ICSE Solutions Mid-point and Its Converse [ Including Intercept Theorem]"},"content":{"rendered":"
ICSE Solutions<\/a>Selina ICSE Solutions<\/a><\/p>\n APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 with Free PDF download option. Selina Publishers Concise Mathematics for Class 9 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.<\/p>\n Download Formulae Handbook For ICSE Class 9 and 10<\/a><\/p>\n Selina ICSE Solutions for Class 9 Maths Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]<\/strong><\/p>\n Exercise 12(A)<\/strong><\/span><\/p>\n Solution 1:<\/strong><\/span> Solution 2:<\/strong><\/span> Solution 3:<\/strong><\/span> Solution 4:<\/strong><\/span> Solution 5:<\/strong><\/span> Solution 6:<\/strong><\/span> Solution 7:<\/strong><\/span> Solution 8:<\/strong><\/span> Solution 9:<\/strong><\/span> Solution 10:<\/strong><\/span> Solution 11:<\/strong><\/span> Solution 12:<\/strong><\/span> Solution 13:<\/strong><\/span> Solution 14:<\/strong><\/span> Solution 15:<\/strong><\/span> Solution 16:<\/strong><\/span> Solution 17:<\/strong><\/span> Exercise 12(B)<\/strong><\/span><\/p>\n Solution 1:<\/strong><\/span> Solution 2:<\/strong><\/span> Solution 3:<\/strong><\/span> Solution 4:<\/strong><\/span> Solution 5:<\/strong><\/span> Solution 6:<\/strong><\/span> Solution 7:<\/strong><\/span> Solution 8:<\/strong><\/span> Solution 9:<\/strong><\/span> Solution 10:<\/strong><\/span> Solution 11:<\/strong><\/span> Solution 12:<\/strong><\/span> Solution 13:<\/strong><\/span> Solution 14:<\/strong><\/span> More Resources for Selina Concise Class 9 ICSE Solutions<\/strong><\/p>\n Selina Concise Mathematics Class 9 ICSE Solutions Mid-point and Its Converse [ Including Intercept Theorem] ICSE SolutionsSelina ICSE Solutions APlusTopper.com provides step by step solutions for Selina Concise Mathematics Class 9 ICSE Solutions Chapter 12 Mid-point and Its Converse [ Including Intercept Theorem]. You can download the Selina Concise Mathematics ICSE Solutions for Class 9 […]<\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_genesis_hide_title":false,"_genesis_hide_breadcrumbs":false,"_genesis_hide_singular_image":false,"_genesis_hide_footer_widgets":false,"_genesis_custom_body_class":"","_genesis_custom_post_class":"","_genesis_layout":"","footnotes":""},"categories":[3034],"tags":[6306,6307,6285,6283,6282,6305,6308],"yoast_head":"\n
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\nGiven ABCD is parallelogram, so AD = BC, AB = CD.
\nConsider triangle APB, given EC is parallel to AP and E is midpoint of side AB. So by midpoint theorem, C has to be the midpoint of BP.
\nSo BP = 2BC, but BC = AD as ABCD is a parallelogram.
\nHence BP = 2AD
\nConsider triangle APB, AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.
\nHence Proved<\/p>\n
\nConsider trapezium ABCD.
\nGiven E and F are midpoints on sides AD and BC, respectively.
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\nConsider LHS,
\nAB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG
\nSo AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF
\nAB + CD = 2EF
\nHence Proved<\/p>\n
\nGiven \u0394 ABC
\nAD is the median. So D is the midpoint of side BC.
\nGiven DE || AB. By the midpoint theorem, E has to be midpoint of AC.
\nSo line joining the vertex and midpoint of the opposite side is always known as median. So BE is also median of \u0394 ABC.<\/p>\n\n