Question

In the given figure, the medians QS and RT of a ∆PQR meet at G. prove that

i) ∆TGS ~ ∆RGQ, and

ii) QG = 2GS from (i) above.​

Answer 1

Heya!!

Here's your answer!!

Statement. Reason

1. S, T are midpoint of PQ (Data)

and PR respectively.

2. TS || QR (Midpoint theorem)

3. TS = 1/2 QR. (Midpoint theorem)

In ∆TGS and ∆RGQ

4. Angle GTS and Angle GRQ. ( Alt angle S TS || QR)

5. Angle GTS and Angle GQR. (Alt Angle S. " " " )

6. Angle SGT and Angle QGR. (vertically opp. angle S )

7. Therefore, ∆SGT = ∆RGQ. (vertically opp. angle S)

8. GS/GQ = TS/QR St. 7

9. GS/GQ = 1/2. St. 3, 7

10. therefore GQ = 2GS. St. 9.

Glad help you,

it helps you

thanks.

Answer 2

(i) QS and RT are medians

Thus, PT/PT = PS/PR, which means that TS || QR

As TS || QR, /_STG = /_GRQ

/_TSG = /_GQR

[Both by alternate interior angles theorem]

Thus, Both the triangles are similar by AA test

(ii) As proved, ∆TSG ~ ∆GQR:

ST/QR = GS/QG = GT/GR

Also, PT/PT = PS/PR = ST/QR = 1/2

[Proved]

=> 1/2 = ST/QR = GS/QG = GT/GR

=> GS/QG = 1/2

=> 2GS = QG

HENCE PROVED.