Question

ABC is an equilateral triangle.
AD is it's altitude. Prove that AD is a median and as well as the bisector of Angle A .

Answer 1

Answer:

Refer to the attachment for the diagram.

Given that,

ABC is equilateral triangle. So the assumptions that can be made are:

• All the angles are equal and measure exactly 60°
• All sides are equal. ( AB = BC = AC )

To Prove:

• AD is the median, i.e. BD = CD ( As medians divide the line into two equal halves )
• ∠ CAD = ∠ BAD

Proof:

Consider Δ ACD and Δ ABD

Here,

∠ ADC = ∠ ADB ( R ) ( 90° each as Height is perpendicular to base )

AC = AB ( H ) ( Equilateral triangle has all sides equal )

AD = AD ( S ) ( Common side )

Hence by RHS congruence rule,

Δ ACD ≅ Δ ABD

By CPCT,

⇒ CD = BD

⇒ ∠ CAD = ∠ BAD

This is the required proof.

Hence Proved !!

Answer 2

According to the question :



• ∆ABC is a equilateral triangle, so
AB = BC = AC


$\rightarrow AB = AC\quad\quad\textit{...(i)}$




• AD is an altitude ( at 90° ) of ∆ABC. It means that a line AD is perpendicular to BC( side opp. to the angle A )


Now,
$\angle ADC = 90° = \angle ADB\quad\quad\textit{...(ii)}$



• AD is a side which is common in both the triangles, therefore
AD of ∆ ADB = AD of ∆ADC



Hence, { From ( i ) , ( ii ) and ( iii ) }
$AD = AD \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( Common ) \\ \angle ADB = \angle ADC \: \: \: \: \: \: \: ( 90° ) \\ AB = AC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ( all \: sides \: are \: equal\: in \: equilateral \: triangle )$



$\triangle ADB \cong \triangle ADC$


By c.p.c.t
BD = CD & /_ BAD = /_ CAD

Hence, AD is a median as well as the bisector of angle A.