Q) In the given figure, CD and RS are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR then prove that:

(i) Δ ADC ~ Δ PSR

(ii) AD x PR = AC x PS

30.1.2_Q34b

Ans:

(i) Its given that:

  1. CD is median of ΔABC
  2. RS is median of ΔPQR
  3. ΔABC∼ΔPQR

AB = 2AD​ as CD is the median

and PQ=2PS​ as RS is the median

Since ΔABC∼ΔPQR

\frac{AB}{PQ} = \frac{AC}{PR}

\frac{2AD}{2PS} = \frac{AC}{PR}

\frac{AD}{PS} = \frac{AC}{PR}

\therefore ΔADC∼ΔPSR ….. Hence proved

(ii)  Since we just proved that, ΔADC∼ΔPSR

\frac{AD}{PS} = \frac{AC}{PR}

\therefore   AD x PR = AC x PS ….. Hence proved

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