In figure, CD and RS are medians of ∆ABC and ∆PQR respectively. If

1.	In figure, CD and RS are medians of ∆ABC and ∆PQR respectively. If ∆ABC ~ ∆PQR then Prove that :(i) ∆ABC ~ ∆PQR(ii) \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)
Answer» (i) Given : Two similar triangles ABC and PQR CD ⊥ AB and RS ⊥ PQ. To prove : ∆ADC ~ ∆PSR Proof : \(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\) ⇒ \(\frac { CA }{ RP }\) = \(\frac { 2AD }{ 2PS }\) (Since D and S. are medians of AB and PQ) Now, in ∆ACD and ∆PRS, \(\frac { CA }{ RP }\) = \(\frac { AD }{ PS }\) and ∠A = ∠P (∵ ∆ABC ~ ∆PQR) ∠ADC = ∠PSR = 90° Thus, by AA similarity Criterion. ∆ADC ~ ∆PSR. (ii) Given : ∆ADC ~ ∆PSR. To prove : \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\) Proof : Given that ∆ABC and ∆PQR are similar. So ∠A = ∠P ….(i) (∵ Corresponding angle S of similar triangles are same.) \(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\) = \(\frac { CB }{ RQ }\) ….(ii) No, In ∆CAD and ∆RPS, ∠A = ∠P (given) ∠CDA = ∠RSP = 90° By A-A similarity criterion ∆CDA ~ ∆RSP \(\frac { CA }{ RP }\) = \(\frac { CD }{ RS }\) From equation (ii) and (iii) \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)

In figure, CD and RS are medians of ∆ABC and ∆PQR respectively. If ∆ABC ~ ∆PQR then Prove that :(i) ∆ABC ~ ∆PQR(ii) \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)

Answer»

(i) Given : Two similar triangles ABC and PQR CD ⊥ AB and RS ⊥ PQ.

To prove : ∆ADC ~ ∆PSR

Proof : \(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\)

⇒ \(\frac { CA }{ RP }\) = \(\frac { 2AD }{ 2PS }\) (Since D and S. are medians of AB and PQ)

Now, in ∆ACD and ∆PRS,

\(\frac { CA }{ RP }\) = \(\frac { AD }{ PS }\)

and ∠A = ∠P (∵ ∆ABC ~ ∆PQR)

∠ADC = ∠PSR = 90°

Thus, by AA similarity Criterion.

∆ADC ~ ∆PSR.

(ii) Given :

∆ADC ~ ∆PSR.

To prove : \(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)

Proof : Given that ∆ABC and ∆PQR are similar.

So ∠A = ∠P ….(i) (∵ Corresponding angle S of similar triangles are same.)

\(\frac { CA }{ RP }\) = \(\frac { AB }{ PQ }\) = \(\frac { CB }{ RQ }\) ….(ii)

No, In ∆CAD and ∆RPS,

∠A = ∠P (given)

∠CDA = ∠RSP = 90°

By A-A similarity criterion

∆CDA ~ ∆RSP

\(\frac { CA }{ RP }\) = \(\frac { CD }{ RS }\)

From equation (ii) and (iii)

\(\frac { CD }{ RS }\) = \(\frac { AB }{ PQ }\)