In a triangle PQR, ST is parallel to QR Show that RT(PQ + PS) = SQ (PR + PT)

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Solution

PQR is a triangle. ST is parallel to QR
By basic proportionality theorem in $Δ P Q R$
$\frac{P Q}{P S} = \frac{P R}{P T}$
By componendo and dividendo
$\frac{P Q}{P S} = \frac{P R}{P T} \Rightarrow \frac{P Q + P S}{P Q - P S} = \frac{P R + P T}{P R - P T}$
$\frac{P Q + P S}{S Q} = \frac{P R + P T}{T R}$
$R T (P Q + P S) = S Q (P R + P T)$

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In a triangle PQR, ST is parallel to QR Show that RT(PQ + PS) = SQ (PR + PT)

PQR is a triangle. ST is parallel to QRBy basic proportionality theorem in ΔPQRPQPS=PRPTBy componendo and dividendoPQPS=PRPT⇒PQ+PSPQ−PS=PR+PTPR−PTPQ+PSSQ=PR+PTTRRT(PQ+PS)=SQ(PR+PT)