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Question
In ▢ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that `"AP"/"PD" = "PC"/"BP"`.
Solution
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 ...(Verticall 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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solution:
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Similarly, in ∆PMR, Ray MY is the bisector of ∠PMR.
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But `("MP")/("MQ") = ("MP")/("MR")` .............(III) [As M is the midpoint of QR.]
Hence MQ = MR
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