Angle Bisector Theorem

In $\text{△ABC}$,$\mathrm{AD}$ is the bisector of $\angle \mathrm{A}$  that intersects BC in D. If $\mathrm{AB} = 8 \mathrm{cm}$ and $\mathrm{AC} = 12 \mathrm{cm},$ then find BD : DC?

In the adjoining figure if exterior $\angle EAB= 110°, \angle CAD= 35°, AB= 5 \mathrm{cm}, AC= 7 \mathrm{cm}\text{ and} BC= 3 \mathrm{cm}$, then $CD$ is equal to:

In a triangle $PQR$, the internal bisectors of $\angle Q\text{ and} \angle R$ meet at $O$. Prove that $OP$ is also internal bisector of $\angle P$. 

In figure if $AB\parallel CD\parallel FE$ then find $x$ and $AE$

The side $BC$ of triangle $ABC$ is produced to $D$.The bisector of angle $A$ meets $BC$ in $L$. Prove that $\angle ABC+\angle ACD=2 \angle ALC$.

In the figure, Point $Q$ is on the side of $MP$ such that $MQ=2$ and $MP=5.5$. Ray $NQ$ is the bisector of $\angle MNP$ of $∆MNP$. If $MN:NP$ is $k:1$,then find $k+1$.

In what ratio of line $x-y-2=0$ divides the line segment joining $\left(3,-1\right)$ and $\left(8,9\right)$?

In $\Delta XYZ, XE$ is the bisector of $\angle X$. Let $XY = 4$ units, $YE = 2$ units and $EZ = 3$ units. Can you find the length of $XZ$( in units)

Amy drew a triangle $ABC$ on the board where $AD$ is the line drawn on side $BC$, where, $AB = 4$ in, $AC = 6$ in, $BD = 1.6$ in and $DC = 2.4$ in. She wants to know whether $AD$ is the angle bisector of $\angle A$. Can you help her?

State and prove the converse of angle bisector theorem

State the exterior angle bisector theorem.

$\mathrm{AE}$ is the bisector of the exterior $\angle \mathrm{CAD}$ meeting $\mathrm{BC}$ produced in $\mathrm{E}$. If $\mathrm{AB}=8 \mathrm{cm}, \mathrm{AC}=4 \mathrm{cm}$ and $\mathrm{BC}=10 \mathrm{cm}$, find $\mathrm{CE}$.

In the given fig.$AM\perp BC$ and $AN$ is the bisector of $\angle A$. If $\angle B=65°$ and $\angle C=33°$, then the value of $\angle MAN$ will be

In the given figure, $AD$ is the bisector of $\angle BAC$. If $AB=10cm, AC=6cm$ and $BC=12cm$. Find $BD$.

In $∆\mathrm{ABC}$, $\angle \mathrm{ABC}=90°$ and$\angle \mathrm{BAC}=60°$. If bisector of $\angle \mathrm{BAC}$ meets $\mathrm{BC}$ at $\mathrm{D}$. Then $\mathrm{BD}:\mathrm{DC}$ is

$D$ and $E$ are the points on sides $BC$ and $AC$, respectively of $∆ABC. AD$ and $BE$ intersect each other at T. If $\frac{AT}{TD}=5$ and $\frac{BT}{ET}=7$ , then $CD : BD =$

In triangle$ABC$ the point$P$ divides the extension of the line $AB$ in the following ratio $AP:BP=AC:BC$. Prove that the line$CP$ is the bisector of the exterior angle $C$.

The external angle bisector of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

In the figure above, $AD\perp AB, CD\perp BC, \angle ABD=x-23°, \angle DBC=1°-x$ and $AD=CD$. Find $x$.

In the figure above, $AD\perp AB, CD\perp BC, \angle ABD=25°$ and $AD=CD$. Find $\angle DBC$.