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

Question

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

Ajanta Trivedi · Accepted Answer

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

TPT: $\frac{P Q^{2}}{P R^{2}} = \frac{P M}{P R}$

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.

$\frac{P Q}{P R} = \frac{P M}{P Q} \Rightarrow P Q^{2} = P M \times P R$

now dividing the whole equation by $P R^{2}$ , we have

$\frac{P Q^{2}}{P R^{2}} = \frac{P M \times P R}{P R^{2}} \frac{P Q^{2}}{P R^{2}} = \frac{P M}{P R}$

which is the required result.

hope this helps you.

cheers!!

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