Question

in triangle pqr angle pqr is equal to 90 degree segment QS is perpendicular to segment QR segment qm is angle bisector of angle pqr prove that p m square upon p r square is equal to PS upon SR

Answer 1

Hope that you were helped

Answer 2

Corresponding Parts of Similar Triangles are Equal

Step-by-step explanation:

In ΔPQR,

Segment QM bisects $\angle PQR$ (Given)

So,

$\frac {PM}{MR} = \frac {PQ}{QR}$ (Using Property of Angle Bisector of Triangle)

Squaring both sides we get-

$\frac {PM^2}{MR^2} = \frac {PQ^2}{QR^2}$ ""(i)

Now, In ΔPQR,

Given,  

$m\angle PQR = 90$

Segment QS is perpendicular to hypotenuse PR

Thus, ΔPQR∼ΔPSQ∼ΔQSR ""(ii) (Right Angled Triangles Similarity Theorem)

ΔPSQ∼ΔPQR

$\frac {PQ}{PR} = \frac {PS}{PQ}$ (Corresponding sides of similar triangles)

So, $PQ^2 = PR \times PS$ ""(iii)

Also, ΔQSR∼ΔPQR ...using (ii)

$\frac {QR}{PR} = \frac {SR}{QR}$ (Corresponding sides of similar triangles)

Therefore, $QR^2 = PR \times SR$ ""(iv)

Hence, $\frac {PM^2}{MR^2} = \frac {PR \times PS}{PR \times SR}$ (From (i),(iii) & (iv))

Finally, $\frac {PM^2}{MR^2} = \frac {PS}{SR}$ PROVED!