Question

Question No. 23
In the given figure, AEDF is a kite with AF = FD and ABCD is a rectangle. If Fis the mid-point of BC, BC = 24 cm, AB = 16 cm and AE = 15 cm, then the perimeter of AFDE is
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Answer 1

Step-by-step explanation:

GIVEN: ABCD is a rectangle, E is the mid point of BC, AE intersects diagonal DB at F

TO PROVE THAT: DF= 2FB, & AF= 2FE

CONSTRUCTION: construct EH || DC, intersecting DA at H.

Since, EH || DC, But DC|| AB

So , EH || AB

Now in TriangleBCD , Since E is the mid point of BC, & EG|| DC, Therefore G should be the mid point of DB, & GE= DC/2 ( By mid point theorem) or, GE= AB/2 ( Since DC= AB)

Now, Since triangle FGE~ Triangle FBA ( BY AA Similarity theorem)

GE/BA = FE/FA ( corresponding sides of similar triangle)

1/2 = FE/ FA

SO, FA= 2FE…………. ( Hence Proved)…

Now, Since triangle FDA ~ Triangle FBE ( by AA similarity theorem)

Therefore, FD/FB = DA/BE ( CSST)

Or, FD/FB = 2/1

So, DF = 2FB……………( Hence proved)

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