Question

in triangle x is a point on side QR of triangle PQR such that PX=PR show that PQ>PX​

Answer 1

Answer:

In Δu PQR  &  Q×R1

QRPR=XRQR      (∵QR2=PR×R)

∡R=∡R

∴    ΔPQR∼ΔQ×R  (SAS criterion)

∴    QXPQ=XRQR=QRPR      (corresponding parts)

⇒QXPQ=VRQR

or,  QXPR=XRQR      (∵PQ=PR)

⇒QR=QX      proved.

I think it would be brainleast answer thanks