In Delta ABC,overline{AD} is a median. If AB | vedclass

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(C) Given that $\overline{AD}$ is a median to side $\overline{BC}$,so $D$ is the midpoint of $\overline{BC}$.Since $AD = BD = 8.5$,in $\Delta ABD$,we

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$15$

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(C) Given that $\overline{AD}$ is a median to side $\overline{BC}$,so $D$ is the midpoint of $\overline{BC}$.Since $AD = BD = 8.5$,in $\Delta ABD$,we have $AD = BD = 8.5$. This implies $\angle DAB = \angle DBA$.In $\Delta ABD$,by the Law of Cosines or using the property of isosceles triangles,we can find $\angle ADB$. Since $AD = BD = 8.5$,$D$ is the center of a circle passing through $A, B,$ and $C$ if $CD = 8.5$ as well.Actually,since $AD$ is a median,$BD = DC = 8.5$. Thus,$AD = BD = DC = 8.5$.This means $D$ is the circumcenter of $\Delta ABC$,and $BC$ is the diameter of the circumcircle.Therefore,$\angle BAC = 90^\circ$ (angle in a semicircle).In right-angled $\Delta ABC$,by Pythagoras theorem,$AC^2 + AB^2 = BC^2$.We know $AB = 8$ and $BC = BD + DC = 8.5 + 8.5 = 17$.$AC^2 + 8^2 = 17^2$$AC^2 + 64 = 289$$AC^2 = 225$$AC = 15$.

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.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 8$ and $AD = BD = 8.5$,find $AC$.

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