Solution - Triangles on the Same Base and Between the Same Parallels



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ConceptTriangles on the Same Base and Between the Same Parallels  


In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i) ar (ACB) = ar (ACF)

(ii) ar (AEDF) = ar (ABCDE)


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In the given figure, E is any point on median AD of a ΔABC. Show that

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In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-


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