Question

class ten exercise 6.3 question 4...... plz helps to solve ..... I have done I just need confirmation ... ​

Answer 1

$\huge\sf\green{solution!!...}$

$\frac{qr}{qs} = \frac{qt}{pr}$

$angle1 = angle2$

$\sf{\red{to\:proof!!...}}$

In ∆PQS = ∆TQR

angle 1 = angle 1 (common)

$\frac{qr}{qs} = \frac{qt}{pr}$ ---- eq {i}

$\sf{\red{now!!...}}$

angle 1 = angle 2 (given)

∆PQR is an isosceles triangle...

PQ = PR ----- eq {ii}

from eq {i} and {ii}...

$\frac{qr}{qs} = \frac{qt}{pq}$

hence, ∆PQS ≈ ∆TQR

❗hope!!...it helps uhh...❗