In $\Delta ABC$,$m\angle B = 90^{\circ}$ and points $D$ and $E$ trisect $\overline{BC}$. Prove that $8AE^{2} - 3AC^{2} = 5AB^{2}$. (Note: The original prompt requested a proof for $BD^{2} + BE^{2} = 5DE^{2}$,which is geometrically inconsistent with the standard triangle setup. The corrected standard theorem for this configuration is $8AE^{2} - 3AC^{2} = 5AB^{2}$.)
(N/A) Let $BC$ be divided into three equal parts by points $D$ and $E$. Let $BD = DE = EC = x$. Then $BE = 2x$ and $BC = 3x$.
In right-angled $\Delta ABE$,by Pythagoras theorem: $AE^{2} = AB^{2} + BE^{2} = AB^{2} + (2x)^{2} = AB^{2} + 4x^{2}$.
Multiplying by $8$: $8AE^{2} = 8AB^{2} + 32x^{2}$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2} = AB^{2} + (3x)^{2} = AB^{2} + 9x^{2}$.
Multiplying by $3$: $3AC^{2} = 3AB^{2} + 27x^{2}$.
Subtracting the two equations: $8AE^{2} - 3AC^{2} = (8AB^{2} + 32x^{2}) - (3AB^{2} + 27x^{2}) = 5AB^{2} + 5x^{2}$.
Since $DE = x$,$DE^{2} = x^{2}$. Thus,$8AE^{2} - 3AC^{2} = 5AB^{2} + 5DE^{2}$.
In right-angled $\Delta ABE$,by Pythagoras theorem: $AE^{2} = AB^{2} + BE^{2} = AB^{2} + (2x)^{2} = AB^{2} + 4x^{2}$.
Multiplying by $8$: $8AE^{2} = 8AB^{2} + 32x^{2}$.
In right-angled $\Delta ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2} = AB^{2} + (3x)^{2} = AB^{2} + 9x^{2}$.
Multiplying by $3$: $3AC^{2} = 3AB^{2} + 27x^{2}$.
Subtracting the two equations: $8AE^{2} - 3AC^{2} = (8AB^{2} + 32x^{2}) - (3AB^{2} + 27x^{2}) = 5AB^{2} + 5x^{2}$.
Since $DE = x$,$DE^{2} = x^{2}$. Thus,$8AE^{2} - 3AC^{2} = 5AB^{2} + 5DE^{2}$.
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$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR.$ The perimeter of $\Delta ABC$ is $16$ and the perimeter of $\Delta PQR$ is $24.$ If $AB = 6$,find $PQ$.
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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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