ABCD is a rectangle. The line through C perpendicular to the diagonal AC intersects AB, AC (both produced) at E and F respectively. Prove that BEFD is a cyclic quadrilateral.
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3
Answer:
Let AB be the chord of the given circle with centre O and a radius of 10 cm.
Then AB =16 cm and OB = 10 cm
From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ BM = (162) cm=8 cm
In the right ΔOMB, we have:
OB2 = OM2 + MB2 (Pythagoras theorem)
⇒ 102 = OM2 + 82
⇒ 100 = OM2 + 64
⇒ OM2 = (100 - 64) = 36
⇒ OM=36−−√ cm=6 cm
Hence, the distance of the chord from the centre is 6 cm.
May be it is help to you
Answered by
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In ΔADE,
∠EAD+∠BAF=90
o
and
∠EAD+∠BAF=90
o
⇒∠EDA=∠BAF
⇒
AF
DE
=
BF
AE
⇒
BF
AF
=
AE
DE
=
3
5
....(1)
similarly ΔCFB:ΔDEC
⇒
BF
CE
=
CF
DE
⇒
BF
CF
=
CE
DE
=
7
5
...(2)
adding (1) and (2)
BF
AF+CF
=
21
50
⇒
BF
AC
=
21
50
⇒
BF
10
=
21
50
⇒BF=4.2
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