In $\Delta PQR$,$\overline{PM}$ is an altitude. If $PM^2 = QM \times RM$,prove that $\Delta PQR$ is a right-angled triangle.
(N/A) Given: In $\Delta PQR$,$\overline{PM} \perp \overline{QR}$ and $PM^2 = QM \times RM$.
Step $1$: From the given equation,we have $\frac{PM}{QM} = \frac{RM}{PM}$.
Step $2$: In $\Delta PMQ$ and $\Delta RMP$,$\angle PMQ = \angle RMP = 90^\circ$ (since $\overline{PM}$ is an altitude).
Step $3$: By the $SAS$ similarity criterion,$\Delta PMQ \sim \Delta RMP$.
Step $4$: Since the triangles are similar,their corresponding angles are equal: $\angle QPM = \angle PRM$ and $\angle PQM = \angle RPM$.
Step $5$: In $\Delta PQR$,the sum of angles is $180^\circ$. Thus,$\angle P + \angle Q + \angle R = 180^\circ$.
Step $6$: Substituting the angles,we get $\angle QPM + \angle RPM + \angle Q + \angle R = 180^\circ$.
Step $7$: Since $\angle QPM = \angle R$ and $\angle RPM = \angle Q$,we have $\angle R + \angle Q + \angle Q + \angle R = 180^\circ$,which simplifies to $2(\angle Q + \angle R) = 180^\circ$,so $\angle Q + \angle R = 90^\circ$.
Step $8$: Therefore,$\angle P = 180^\circ - 90^\circ = 90^\circ$. Hence,$\Delta PQR$ is a right-angled triangle.
Step $1$: From the given equation,we have $\frac{PM}{QM} = \frac{RM}{PM}$.
Step $2$: In $\Delta PMQ$ and $\Delta RMP$,$\angle PMQ = \angle RMP = 90^\circ$ (since $\overline{PM}$ is an altitude).
Step $3$: By the $SAS$ similarity criterion,$\Delta PMQ \sim \Delta RMP$.
Step $4$: Since the triangles are similar,their corresponding angles are equal: $\angle QPM = \angle PRM$ and $\angle PQM = \angle RPM$.
Step $5$: In $\Delta PQR$,the sum of angles is $180^\circ$. Thus,$\angle P + \angle Q + \angle R = 180^\circ$.
Step $6$: Substituting the angles,we get $\angle QPM + \angle RPM + \angle Q + \angle R = 180^\circ$.
Step $7$: Since $\angle QPM = \angle R$ and $\angle RPM = \angle Q$,we have $\angle R + \angle Q + \angle Q + \angle R = 180^\circ$,which simplifies to $2(\angle Q + \angle R) = 180^\circ$,so $\angle Q + \angle R = 90^\circ$.
Step $8$: Therefore,$\angle P = 180^\circ - 90^\circ = 90^\circ$. Hence,$\Delta PQR$ is a right-angled triangle.
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View SolutionWhich of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$ $a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity. $2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$ $b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity. $3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$ $c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity. $4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$ $d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.
| Part $I$ | Part $II$ |
| $1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$ | $a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity. |
| $2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$ | $b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity. |
| $3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$ | $c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity. |
| $4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$ | $d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity. |
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