As shown in the adjoining diagram,$\angle B$ is an obtuse angle and $\overline{AM}$ is an altitude in $\Delta ABC$. Prove that $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.
(N/A) In $\Delta AMC$,$\angle AMC = 90^{\circ}$. By Pythagoras theorem,$AC^{2} = AM^{2} + MC^{2}$.
Since $MC = MB + BC$,we have $AC^{2} = AM^{2} + (MB + BC)^{2}$.
Expanding the square,$AC^{2} = AM^{2} + MB^{2} + BC^{2} + 2 \cdot MB \cdot BC$.
In $\Delta AMB$,$\angle AMB = 90^{\circ}$. By Pythagoras theorem,$AB^{2} = AM^{2} + MB^{2}$.
Substituting $AB^{2}$ for $AM^{2} + MB^{2}$ in the equation,we get $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.
Since $MC = MB + BC$,we have $AC^{2} = AM^{2} + (MB + BC)^{2}$.
Expanding the square,$AC^{2} = AM^{2} + MB^{2} + BC^{2} + 2 \cdot MB \cdot BC$.
In $\Delta AMB$,$\angle AMB = 90^{\circ}$. By Pythagoras theorem,$AB^{2} = AM^{2} + MB^{2}$.
Substituting $AB^{2}$ for $AM^{2} + MB^{2}$ in the equation,we get $AC^{2} = AB^{2} + BC^{2} + 2 \cdot BC \cdot BM$.
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In parallelogram $ABCD$,$A-P-B$ and $AP = \frac{2}{3} AB$. $\overline{DP}$ intersects $\overline{AC}$ at $Q$. Find the ratio of the areas of $\Delta APQ$ and $\Delta CDQ$.
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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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View Solution............... is not a condition for the similarity of two triangles.
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