(D) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportio

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(D) According to the Angle Bisector Theorem,if a ray bisects an angle of a triangle,then it divides the opposite side into segments that are proportional to the other two sides of the triangle.In $\Delta ABC$,since $AD$ is the bisector of $\angle A$,it divides the side $BC$ such that:$\frac{BD}{DC} = \frac{AB}{AC}$By cross-multiplying the terms,we get:$BD \times AC = DC \times AB$Therefore,the correct option is $D$.

Class 10 Mathematics Triangles Mix Examples - Triangles

Medium

In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at point $D$,then

A
$BD \times AC = BC \times AB$
B
$BD \times AB = DC \times AC$
C
$AC \times AB = DC \times BC$
D
$BD \times AC = DC \times AB$

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is a median. If $AB = 5$ and $BC = 12$,find $BD$.

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In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $\frac{AB}{CD} = \frac{2}{1}$ and $AC = 15$,find $MA$ and $MC$.

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$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$. If $PQ = 8$,$XZ = 12$ and the area of $\Delta PQR$ is $24$,find the area of $\Delta XYZ$.

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In $\Delta ABC$,$AB > AC$ and $D$ is the midpoint of $\overline{BC}$. $\overline{AM} \perp \overline{BC}$ and $M \in \overline{BC}$. Prove that $AB^{2} - AC^{2} = 2 \cdot BC \cdot DM$.

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BE}$ is an altitude to the hypotenuse $\overline{AC}$. Prove that $\frac{AB^2}{BC^2} = \frac{AE}{CE}$.

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In $\Delta ABC$,the bisector of $\angle A$ intersects $\overline{BC}$ at point $D$,then

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BD}$ is a median. If $AB = 5$ and $BC = 12$,find $BD$.

In $\square ABCD$,$\overline{AB} \parallel \overline{CD}$ and $\overline{AC} \cap \overline{BD} = \{M\}$. If $\frac{AB}{CD} = \frac{2}{1}$ and $AC = 15$,find $MA$ and $MC$.

$\Delta PQR \sim \Delta XYZ$ for the correspondence $PQR \leftrightarrow XZY$. If $PQ = 8$,$XZ = 12$ and the area of $\Delta PQR$ is $24$,find the area of $\Delta XYZ$.

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In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BE}$ is an altitude to the hypotenuse $\overline{AC}$. Prove that $\frac{AB^2}{BC^2} = \frac{AE}{CE}$.

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