Question
$\mathrm{PQR}$ is a triangle right angled at $\mathrm{P}$ and $\mathrm{M}$ is a point on QR such that PM $\perp$ QR. Show that $\mathrm{PM}^{2}=\mathrm{QM} . \mathrm{MR}$
Step 1
We also mark the point M on QR such that PM is perpendicular to QR. Show more…
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Sides $\mathrm{AB}$ and $\mathrm{BC}$ and median $\mathrm{AD}$ of $\mathrm{a}$ triangle $\mathrm{ABC}$ are respectively proportional to sides $\mathrm{PQ}$ and $\mathrm{QR}$ and median PM of $\Delta$ PQR (see Fig. $6.41$ ). Show that $\Delta \mathrm{ABC}-\Delta \mathrm{PQR}$
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In Fig. $6.39, \mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively. Prove that: (i) $\Delta \mathrm{ABC}-\Delta \mathrm{AMP}$ (ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
$\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\Delta$ PQR such that $\angle P=\angle$ RTS. Show that $\Delta \mathrm{RPQ}-\Delta \mathrm{RTS}$.
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