In a $\triangle\text{ABC},\text{M}$ and N are points on the sides AB and AC respectively such that BM || BC.
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In $\triangle\text{ABC},\angle\text{B}=\angle\text{C}$
$\therefore\text{AB}=\text{AC}$ (Sides opposite to equal angle are equal)
Subtracting BM from both sides, we get
AB - BM = AC - BM
⇒ AB - BM = AC - CN $(\therefore\text{BM=CN})$
⇒ AM = AN
$\therefore\angle\text{AMN}=\angle\text{ANM}$ (Angels opposite to equal sides are equal)
Now, in $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\dots(1)$(Angle Sum Property of triangle)
Again in $\triangle\text{AMN},$
$\angle\text{A}+\angle\text{AMN}+\angle\text{ANM}=180^\circ\dots(2)$(Angle Sum Property of triangle)
From (1) and (2), we get
$\angle\text{B}+\angle\text{C}=\angle\text{AMN}+\angle\text{ANM}$
$\Rightarrow2\angle\text{B}=2\angle\text{ANM}$
$\Rightarrow\angle\text{B}=\angle\text{AMN}$
Since, $\angle\text{B}$ and $\angle\text{AMN}$ are corresponding angles.
$\therefore\text{MN }||\text{ BC}$
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For the following statments state whether true (T) or false(F):
The ratio of the areas of two similar triangles is equal to the ratio of their corresponding angle-bisector segments.
In the given figure, side BC of $\triangle\text{ABC}$ is bisected at D and O is any point on AD. BO and CO produced meet AC and AB at E and F respectively, and AD is produced to X so that D is the midpoint of OX Prove that AO : AX = AF : AB and show that EF || BC.
In the given figure, $\angle\text{AMN}=\angle\text{MBC}=76^\circ.$ If p, q and r are the lengths of AM, MB and BC respectively then express the length of MN in terms of p, q and r.