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Question
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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Solution
Given, In ΔACB,
m∠ACB = 90°
∵ seg CD ⊥ hypo AB
∴ seg CE is angle bisector ∠ACB.
To prove that: `(AD)/(BD) = (AE^2)/(BE^2)`
Proof: In ΔABC
seg CE is angle bisector of ∠ACB. ......(Given)
∴ `(AC)/(BC) = (AE)/(BE)` ......[Angle bisector theorem]
Squaring on both the sides
`(AC^2)/(BC^2) = (AE^2)/(BE^2)` ......(i)
In ΔACB , m∠ACB = 90°
And seg CD ⊥ hypotenuse AB ......(Given)
∴ CD2 = AD × BD ......[Geometric mean thorem]
Dividing both the sides by BD2
`(CD^2)/(BD^2) = (AD xx BD)/(BD^2)`
`(CD^2)/(BD^2) = (AD)/(BD)` .......(ii)
Now, in the figure
ΔACB ∼ ΔADC ∼ ΔCDB ......(Right-angled triangle similarity property)
Consider,
ΔADC ∼ ΔCDB
`(AC)/(BC) = (DC)/(BD)` ......(c.s.c.t)
`(AC^2)/(BC^2) = (DC^2)/(BD^2)`
∴ `(AC^2)/(BC^2) = (AE^2)/(BE^2) = (AD)/(BD)` .......[From equation (i), (ii) and (iii)]
∴ `(AD)/(BD) = (AE^2)/(BE^2)`
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In ΔABC, ray BD bisects ∠ABC.
If A – D – C, A – E – B and seg ED || side BC, then prove that:
`("AB")/("BC") = ("AE")/("EB")`
Proof :
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 the figure, ray YM is the bisector of ∠XYZ, where seg XY ≅ seg YZ, find the relation between XM and MZ.
Draw seg AB = 6.8 cm and draw perpendicular bisector of it.
In the following figure, ray PT is the bisector of ∠QPR Find the value of x and perimeter of ∠QPR.
Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.
If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?
In ∆PQR seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR.
Complete the proof by filling in the boxes.
solution:
In ∆PMQ,
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 :
In ΔABC, ray BD bisects ∠B.
∴ `square/("BC") = ("AD")/("DC")` ...(I) (`square`)
ΔABC, DE || BC
∴ `(square)/("EB") = ("AD")/("DC")` ...(II) (`square`)
∴ `("AB")/square = square/("EB")` ...[from (I) and (II)]
