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# As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t

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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
Where are the assumption provided that FC=EC=BC–BE=BC–BD, is it a rule .
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As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
Hello Moderators,

Please confirm if my assumption and understanding to the solution provided by Math Revolution are correct.

Concept - Length of the two tangents, only two possible, drawn from the same point are equal.

Points from which the tangents are drawn are A,B,C

Tangents from A : - AD and AF ; AD = AF = 3

Tangents from B :- BD and BE ; BD = BE = 5

Tangents from C :- CE and CF; CE = CF = 4 (BC = 9 ; BE = 5; EC = FC = (9-5 = 4)

Total length of AC = AF+FC = 3+4 = 7

Thanks

Originally posted by adizephyr on 15 Feb 2020, 08:54.
Last edited by adizephyr on 15 Feb 2020, 09:18, edited 1 time in total.
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
DUtkarsh wrote:
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BI2 + DI2 = BI2 = BE2 + EI2
BI2 + DI2 = BE2 + EI2
BI2 = BE2 (Since, DI = EI = r)
Hence, BI = BE = 5
Therefore, EC = 9 – 5 = 4

Similarly, using triangles ADI and AFI
And, using triangles CEI and CFI
EC = CF = 4

Solution = AF + CF = 3 + 4 = 7 i.e. Option E.

DUtkarsh

can u please elaborate above highlighted portion....
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
1
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Rohit2015 wrote:
DUtkarsh wrote:
Considering the radius of Incircle as r. Know that DI, EI and FI are perpendiculars on their respective sides.
In triangles BDI and BEI,
BI2 + DI2 = BI2 = BE2 + EI2
BI2 + DI2 = BE2 + EI2
BI2 = BE2 (Since, DI = EI = r)
Hence, BI = BE = 5
Therefore, EC = 9 – 5 = 4

Similarly, using triangles ADI and AFI
And, using triangles CEI and CFI
EC = CF = 4

Solution = AF + CF = 3 + 4 = 7 i.e. Option E.

DUtkarsh

can u please elaborate above highlighted portion....

Rohit2015 - Thanks for pointing out the typo, it'd be BD and not BI in the first half of the equation. I hope it is clear now, and I have also updated the aforementioned post.
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Re: As the figure shows AD = 3, BD = 5 and BC = 9. Moreover, point I is t [#permalink]
MathRevolution wrote:
=>

Attachment:
2.3ps(a).png

Since $$AD = AF,$$ we have $$AF = 3.$$

Since $$BC = 9$$ and $$BD = 5,$$ we have $$FC = EC = BC – BE = BC – BD = 9 – 5 = 4.$$

Then we have $$AC = AF + FC = 3 + 4 = 7.$$

Hello MathRevolution,