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In a Parallelogram Abcd, Ab = 10 Cm, Ad = 6 Cm. the Bisector of ∠A Meets Dc in E, Aeand Bc Produced Meet at F. Find Te Length Cf. - Mathematics

In a parallelogram ABCDAB = 10 cm, AD = 6 cm. The bisector of ∠A meets DC in EAEand BC produced meet at F. Find te length CF

 

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Solution

\[\text{ AE is the bisector of } \angle A . \]
\[ \therefore \angle DAE = \angle BAE = x\]
\[ \angle BAE = \angle AED = x (\text{ alternate angles })\]
\[\text{ Since opposite angles in ∆ ADE are equal, ∆ ADE is an isosceles triangle } . \]
\[ \therefore AD = DE = 6 cm (\text{ sides opposite to equal angles })\]
\[AB = CD = 10 cm \]
\[CD = DE + EC\]
\[ \Rightarrow EC = CD - DE\]
\[ \Rightarrow EC = 10 - 6 = 4 cm\]
\[\angle DEA = \angle CEF = x (\text{ vertically opposite angle })\]
\[\angle EAD = \angle EFC = x (\text{ alternate angles })\]
\[\text{ Since opposite angles in } ∆ EFC \text{ are equal, ∆ EFC is an isosceles triangle } . \]
\[ \therefore CF = CE = 4 \text{ cm (sides opposite to equal angles })\]
\[ \therefore CF = 4\text{ cm }\]

 

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APPEARS IN

RD Sharma Class 8 Maths
Chapter 17 Understanding Shapes-III (Special Types of Quadrilaterals)
Exercise 17.1 | Q 29 | Page 12
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