In a trapezium ABCD, AB||CD and DC=3AB. EF||AB intersects DA and CB respectively at E and F such that BF/FC=2/3. Prove that 3DC=5EF
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Step-by-step explanation:
(c): In AAJD and AAGE, ZGAE - ZDAJ [Common]
ZAGE = ZAJD AAJD - AAGE [Each 90°)
[By AA similarity] AG GE = AJ JD
Given, BF BF FC 3
=
In ΔΒΓ and BIC
ZHBF = ZCBI ZBHF = ZBIC
ABHF - ABIC HF BH BE
2
..(i)
[Common] [Each 90° [By AA similarity]
..(ii)
ICBC 5 Now, in rectangles ABHG and ABI) corresponding
sides are parallel and internal angles are 90°.
AG BH
= AJ BI
From (i), (ii) and (iii), we get GE HF 2
JD IC » JD 5 GE 2 5. GE and IC = HF
Now, DC = DJ + JI+ IC
5
2
5
GE + AB + HF
2
ЗАВ
2AB = 4 (GE + HF) 2
5 AB = GE + HF Also, EF = EG + GH + HF
4 5 AB + AB = 9 5 AB
5EF = 9AB = 9 19 (4DC) -
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