In figure below, $Δ\ ABC$ is a triangle such that $\frac{AB}{AC}\ =\ \frac{BD}{DC}$, $∠B\ =\ 70^o$, $∠C\ =\ 50^o$, find $∠BAD$.
"
Given:
In the given figure, $Δ\ ABC$ is a triangle such that $\frac{AB}{AC}\ =\ \frac{BD}{DC}$, $∠B\ =\ 70^o$ and $∠C\ =\ 50^o$.
To do:
We have to find the measure of $∠BAD$.
Solution:
In $\vartriangle ABC$,
$\angle A+\angle B+\angle C=180^o $ [Angle sum property of a triangle]
$\angle A=180^o-(70+50)^o$
$\angle A=60^o$
$\frac{AB}{AC} = \frac{BD}{DC}$ (given)
Therefore, AD is the angle bisector of $\angle A$.
This implies,
$\angle BAD=\frac{\angle A}{2} =\left(\frac{60}{2}\right)^o =30^o $
The measure of $\angle BAD$ is $30^o $.
- Related Articles
- In figure below, $∠\ ABC\ =\ 90^o$ and $BD\ ⊥\ AC$. If $BD\ =\ 8\ cm$, and $AD\ =\ 4\ cm$, find $CD$."\n
- In figure below, $∠\ ABC\ =\ 90^o$ and $BD\ ⊥\ AC$. If $AC\ =\ 5.7\ cm$, $BD\ =\ 3.8\ cm$ and $CD\ =\ 5.4\ cm$, find $BC$."\n
- In figure below, $∠A\ =\ ∠CED$, prove that $ΔCAB\ ∼\ ΔCED$. Also find the value of $x$."\n
- ABCD is a trapezium in which AB ll CD and AD = BC. Show that:(i) ∠A = ∠B (ii) ∠C = ∠D(iii) ∆ABC ≅ ∆BAD(iv) Diagonal AC = Diagonal BD"
- If $Δ\ ABC$ and $Δ\ AMP$ are two right triangles, right angled at $B$ and $M$, respectively such that $∠MAP\ =\ ∠BAC$. Prove that:(i) $ΔABC\ ∼\ ΔAMP$(ii) $\frac{CA}{PA}\ =\ \frac{BC}{MP}$"\n
- In figure, if $TP$ and $TQ$ are the two tangents to a circle with centre $O$ so that $∠POQ = 110^o$, then $∠PTQ$ is equal to(a) $60^o$(b) $70^o$(c) $80^o$(d) $90^o$"
- In a quadrilateral, CO and DO are the bisectors of ∠C and ∠D respectively.Prove that ∠COD=12(∠A + ∠B)."\n
- Prove that ∠ACP = ∠QCD."\n
- In $\triangle ABC$, if $\angle 1=\angle 2$, prove that $\frac{AB}{AC}=\frac{BD}{DC}$."\n
- In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.If $BD\ =\ 2.5\ cm$, $AB\ =\ 5\ cm$, and $AC\ =\ 4.2\ cm$, find $DC$."\n
- In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$.If $BD\ =\ 2\ cm$, $AB\ =\ 5\ cm$, and $DC\ =\ 3\ cm$, find $AC$."\n
- In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$. If $AB\ =\ 3.5\ cm$, $AC\ =\ 4.2\ cm$, and $DC\ =\ 2.8\ cm$, find $BD$. "\n
- $ABC$ is a triangle in which $∠\ A\ =\ 90^o$, $AN\ ⊥\ BC$, $BC\ =\ 12\ cm$ and $AC\ =\ 5\ cm$. Find the ratio of the areas of $ΔANC$ and $ΔABC$. "\n
- In given figure, ABCD is a rhombus in which angle ABD = 40o.Find:i. ∠BACii. ∠BDCiii. ∠ACD"\n
- In a $Δ\ ABC$, $AD$ is the bisector of $∠\ A$, meeting side $BC$ at $D$. If $AB\ =\ 10\ cm$, $AC\ =\ 14\ cm$, and $BC\ =\ 6\ cm$, find $BD$ and $DC$. "\n
Kickstart Your Career
Get certified by completing the course
Get Started