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Question
In ∆ABC, seg BD bisects ∠ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.
Solution
In △ABC,
seg BD bisects ∠ABC. ...(Given)
∴ by the theorem of angle bisector of a triangle,
∴ `"AB"/"BC" = "AD"/"DC"`
∴ `x/(x + 5) = (x – 2)/(x + 2)`
∴ x(x + 2)= (x – 2)(x + 5)
∴ x2 + 2x = x(x + 5) - 2(x + 5)
∴ x2 + 2x = x2 + 5x - 2x - 10
∴ x2 + 2x = x2 + 3x - 10
∴ x2 - x2 + 2x - 3x = - 10
∴ - x = - 10
∴ x = 10
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In ΔABC, ray BD bisects ∠ABC.
∴ `("AB")/("BC") = (......)/(......)` ......(i) (By angle bisector theorem)
In ΔABC, seg DE || side BC
∴ `("AE")/("EB") = ("AD")/("DC")` ....(ii) `square`
∴ `("AB")/square = square/("EB")` [from (i) and (ii)]
In ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.
Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.
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Complete the proof by filling in the boxes.
solution:
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Ray MX is the bisector of ∠PMQ.
∴ `("MP")/("MQ") = square/square` .............(I) [Theorem of angle bisector]
Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
∴ `("MP")/("MR") = square/square` .............(II) [Theorem of angle bisector]
But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
∴ `("PX")/square = square/("YR")` .............[From (I), (II) and (III)]
∴ XY || QR .............[Converse of basic proportionality theorem]
In ΔABC, ray BD bisects ∠ABC, A – D – C, seg DE || side BC, A – E – B, then for showing `("AB")/("BC") = ("AE")/("EB")`, complete the following activity:
Proof :
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∴ `("AB")/square = square/("EB")` ...[from (I) and (II)]
