in triangle PQN, PQ=PN, P-M-N and M-N-R

seg QM and QR are drawn such that angle NQM= angle NQR

prove that: PQ2/PR2=PM/PR

given: in triangle PQN, PQ=PN, and QM and QN are drawn such that ∠NQM=∠NQR,



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]


now in the triangle QMR,




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.


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