In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$. If $\angle B = 60^o$ and $\angle A = 70^o$, prove that $AD > AC$.
Given:
In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$.
$\angle B = 60^o$ and $\angle A = 70^o$.
To do:
We have to prove that $AD > CD$.
Solution:
In $\triangle ABC$,
$\angle CBD + \angle CBA = 180^o$ (Linear pair)
$\angle CBD + 60^o = 180^o$
$\angle CBD = 180^o - 60^o = 120^o$
In $\triangle BCD$,
$BD = BC$
$\angle CDB = \angle BCD$ (Angles opposite to equal sides are equal)
$\angle CDB + \angle BCD = 180^o - 120^o = 60^o$
$\angle CDB = \angle BCD = \frac{60^o}{2} = 30^o$
In $\triangle ABC$,
$\angle A + \angle B + \angle C = 180^o$
$70^o + 60^o + \angle C = 180^o$
$\angle C = 180^o-130^o$
$\angle C = 50^o$
This implies,
$\angle ACD = \angle ACB + \angle BCD$
$ = 50^o + 30^o$
$= 80^o$
$\angle ACD = 80^o$ and $\angle D = 30^o$
Therefore,
$AD > AC$
Hence proved.
Related Articles
- In $\triangle ABC$, side $AB$ is produced to $D$ so that $BD = BC$. If $\angle B = 60^o$ and $\angle A = 70^o$, prove that $AD > CD$.
- In a $\triangle ABC$, if $AB = AC$ and $\angle B = 70^o$, find $\angle A$.
- In $\triangle ABC, BD \perp AC$ and $CE \perp AB$. If $BD$ and $CE$ intersect at $O$, prove that $\angle BOC = 180^o-\angle A$.
- In the given figure, $ABC$ is triangle in which $\angle ABC > 90^o$ and $AD \perp CB$ produced. Prove that $AC^2 = AB^2 + BC^2 + 2BC \times BD$"
- $ABC$ is a triangle in which $\angle B = 2\angle C, D$ is a point on $BC$ such that $AD$ bisects $\angle BAC$ and $AB = CD$. Prove that $\angle BAC = 72^o$.
- In the given figure, $ABC$ is a triangle in which $\angle ABC = 90^o$ and $AD \perp CB$. Prove that $AC^2 = AB^2 + BC^2 - 2BC \times BD$"
- In a $\triangle ABC$, it is given that $AB = AC$ and the bisectors of $\angle B$ and $\angle C$ intersect at $O$. If $M$ is a point on $BO$ produced, prove that $\angle MOC = \angle ABC$.
- In a $\triangle ABC$, if $\angle A = 120^o$ and $AB = AC$. Find $\angle B$ and $\angle C$.
- In a $\triangle ABC, \angle ABC = \angle ACB$ and the bisectors of $\angle ABC$ and $\angle ACB$ intersect at $O$ such that $\angle BOC = 120^o$. Show that $\angle A = \angle B = \angle C = 60^o$.
- In a $\triangle ABC, \angle A = x^o, \angle B = 3x^o and \angle C = y^o$. If $3y - 5x = 30$, prove that the triangle is right angled.
- $BD$ and $CE$ are bisectors of $\angle B$ and $\angle C$ of an isosceles $\triangle ABC$ with $AB = AC$. Prove that $BD = CE$.
- Construct a $∆ABC$ in which $BC = 3.6\ cm, AB + AC = 4.8\ cm$ and $\angle B = 60^o$.
- $O$ is the circumcentre of the triangle $ABC$ and $OD$ is perpendicular on $BC$. Prove that $\angle BOD = \angle A$.
- Prove that each angle of an equilateral triangle is $60^o$.
Kickstart Your Career
Get certified by completing the course
Get Started