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Question

In a parallelogram ABCD, from vertex D, a line is draw which intersects produced BA and BC at E and F respectively. prove ADAE=EBBE=FCCD
1051351_8a0bad24ee69479aa34d9de931e2e234.png

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Solution

In EAD and DCF
1=2[ABCD]
3=4[ADBC]
EADDCF
[By AA similarity]
EADC=ADCF=DEFD [BySSS]
ADAE=CFCD(2)
In EAD and EBF
1=1 (common
3=4[ADBC]
EADEBF [By AA similarity]
EAEB=ADBF=DEEF [By SSS]
ADAE=FBBE(2)
From (1) and (2) , ADAE=FBBE=CFCD

969388_1051351_ans_d0ff612dc4204efdba1e1c69dd2246b6.png

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