Question

In two triangles ABC and PQR, if AD and PS are medians to ∆ABC and ∆PQR respectively and ∆ABD~∆PQS, then prove that ∆ABC~∆PQR​

Answer 1

Answer:

Given data: AD and PS are the two medians drawn to two triangles ∆ABC and ∆PQR respectively. Also, ∆ABD~∆PQS.

To prove: ∆ABC~∆PQR

Now, ∆ABD~∆PQS

∴ AB/PQ = BD/QS = AD/PS …. [ corresponding sides of similar triangle are proportional]

Also, ∠B = ∠Q

Since, sides opposite to equal angles are also equal. So, side opposite to ∠B i.e., AC is equal to side opposite to ∠Q i.e., PR.

Therefore, we can say  

AB/PQ = BD/QS = AD/PS = AC/PR …. (i)

AD is the median on the side BC of ∆ABC

∴BD = DC = ½ BC …. (ii)

Similarly, PS is the median on the side QR of ∆PQR

∴QS = SR = ½ QR …. (iii)

From Equation (i), (ii) & (iii), we can now say

BD/QS = DC/SR = BC/QR …… (iv)

From Equation (i) & (iv), we get

AB/PQ = AC/PR = BC/QR

Also, ∠B = ∠Q

by SSS or SAS property

∴ ∆ABC~∆PQR

Hence proved