# in triangle PQN, PQ=PN, P-M-N and M-N-Rseg QM and QR are drawn such that angle NQM= angle NQRprove that: PQ2/PR2=PM/PR given: in triangle PQN, PQ=PN, and QM and QN are drawn such that ∠NQM=∠NQR,

TPT: proof:

let ∠NQM=∠NQR=α

let ∠PQN=∠PNQ=θ  [since PQ=PN angles opposite to equal sides are equal]

∠QMP=θ+α.........(1) [exterior angle is equal to the sum of two opposite angles, where QMN is the triangle and ∠QMP is the exterior angle]

∠QMR=180-∠QMP [angles formed at one side of straight line are supplementary]

∠QMR=180-(θ+α)

now in the triangle QMR,

∠QRM=180-(2α+180-(θ+α))

∠QRM=θ-α.......(2)

∠PQM=∠PQN-∠MQN=θ-α..........(3)

now in the triangle PQM and triangle PRQ,

∠QPM=∠QPR [same angle]

∠PQM=∠QRP  [by (2) and (3)and QRM is same as QRP]

therefore the triangle PQM and triangle PRQ are similar triangle.

thus the ratios of the corresponding sides are equal. now dividing the whole equation by , we have which is the required result.

hope this helps you.

cheers!!

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