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प्रश्न
In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that `"AP"/"PD" = "PC"/"BP"`.
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उत्तर
Given: ▢ABCD is a parallelogram. seg AD || seg BC and BD is their transversal.
To prove: `"AP"/"PD" = "PC"/"BP"`
Proof:
seg AD || seg BC and BD is their transversal. ...(Given)
∴ ∠DBC ≅ ∠BDA ...(Alternate angles)
∴ ∠PBC ≅ ∠PDA ...(i) [D−P−B]
In △PBC and △PDA,
∠PBC ≅ ∠PDA ...[From (i)]
∠BPC ≅ ∠DPA ...(Vertically opposite angles)
By AA test of similarity,
∴ △APD ∼ CPB
∴ `"AP"/"PC" = "PD"/"PB"` ...(Corresponding sides of similar triangles)
∴ `"AP"/"PD" = "PC"/"PB"` ...(By alternendo)
Hence proved.
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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 ΔABC, ∠ACB = 90°. seg CD ⊥ side AB and seg CE is angle bisector of ∠ACB.
Prove that: `(AD)/(BD) = (AE^2)/(BE^2)`.
Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.
From the information given in the figure, determine whether MP is the bisector of ∠KMN.
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)]
