In an equilateral $\Delta ABC,AD\bot BC$. Prove that $A{{B}^{2}}+D{{C}^{2}}=\dfrac{5}{4}A{{C}^{2}}.$
Answer
362.7k+ views
Hint: Use the property that the altitude divides the triangle into two equal halves. Then use the Pythagoras theorem to both the triangles and hence find the relation between equations to get the desired results.
Complete step-by-step answer:
In the given question we are given an equilateral $\Delta ABC$ where a point D is specified on BC such that $AD\bot BC.$
So this means in the figure below, $\angle ADC=\angle ADB={{90}^{\circ }}$
At first we will write the properties of equilateral triangle which are as follows:
i) All the sides are equal.
ii) All the three internal angles are ${{60}^{\circ }}$ .
iii) The altitude, median, angle bisector and perpendicular bisector for each side are all in the same line.
Thus by the property (iii) of an equilateral triangle we can say that the point ‘D’ divides the sides BC into two equal halves.
So we can say that BD = DC.
As AD divides the equilateral $\Delta ABC$ in two congruent right angled triangles we can apply Pythagora's theorem to both parts of equilateral triangle.
So now by using Pythagoras theorem in $\Delta ADB$ we can say,
$A{{D}^{2}}+B{{D}^{2}}=A{{B}^{2}}..........\left( i \right)$
Similarly, by using Pythagoras theorem in $\Delta ADC$, we can say,
$A{{D}^{2}}+D{{C}^{2}}=A{{C}^{2}}..........\left( ii \right)$
BY moving $A{{D}^{2}}$ from left hand to right hand side in equation (ii) we can write it as,
$D{{C}^{2}}=A{{C}^{2}}-A{{D}^{2}}..........\left( iii \right)$
The equation (i) can be written as,
$A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}..........\left( iv \right)$
So now adding (iii) and (iv) we get,
$A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+B{{D}^{2}}..........\left( v \right)$
Now by the property of equilateral triangle, that is all the sides are equal, we can say that,
AC = BC ………………(vi)
As we know that point ‘D’ divides ‘BC’ into two equal parts so we can write,
BC = 2BD ……………(vii)
Now in equation (vi) we can substitute the result of (vii) so now we can write it as,
AC = 2BD
Which can be further written as,
$BD=\dfrac{AC}{2}$
Now squaring on both the sides we get,
$B{{D}^{2}}=\dfrac{A{{C}^{2}}}{4}$
Now putting $B{{D}^{2}}=\dfrac{A{{C}^{2}}}{4}$ in equation (v) we get,
$\begin{align}
& A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+B{{D}^{2}} \\
& \Rightarrow A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+\dfrac{A{{C}^{2}}}{4} \\
\end{align}$
Hence we can write,
$A{{B}^{2}}+D{{C}^{2}}=\dfrac{5}{4}A{{C}^{2}}$
Hence proved
Note: Students generally stuck after reading to the equation $A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+B{{D}^{2}}.$
They generally forget to use the property that the altitude, median and angle bisector are aligned in the same line. So they should be careful about it.
Complete step-by-step answer:
In the given question we are given an equilateral $\Delta ABC$ where a point D is specified on BC such that $AD\bot BC.$
So this means in the figure below, $\angle ADC=\angle ADB={{90}^{\circ }}$

At first we will write the properties of equilateral triangle which are as follows:
i) All the sides are equal.
ii) All the three internal angles are ${{60}^{\circ }}$ .
iii) The altitude, median, angle bisector and perpendicular bisector for each side are all in the same line.
Thus by the property (iii) of an equilateral triangle we can say that the point ‘D’ divides the sides BC into two equal halves.
So we can say that BD = DC.
As AD divides the equilateral $\Delta ABC$ in two congruent right angled triangles we can apply Pythagora's theorem to both parts of equilateral triangle.
So now by using Pythagoras theorem in $\Delta ADB$ we can say,
$A{{D}^{2}}+B{{D}^{2}}=A{{B}^{2}}..........\left( i \right)$
Similarly, by using Pythagoras theorem in $\Delta ADC$, we can say,
$A{{D}^{2}}+D{{C}^{2}}=A{{C}^{2}}..........\left( ii \right)$
BY moving $A{{D}^{2}}$ from left hand to right hand side in equation (ii) we can write it as,
$D{{C}^{2}}=A{{C}^{2}}-A{{D}^{2}}..........\left( iii \right)$
The equation (i) can be written as,
$A{{B}^{2}}=A{{D}^{2}}+B{{D}^{2}}..........\left( iv \right)$
So now adding (iii) and (iv) we get,
$A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+B{{D}^{2}}..........\left( v \right)$
Now by the property of equilateral triangle, that is all the sides are equal, we can say that,
AC = BC ………………(vi)
As we know that point ‘D’ divides ‘BC’ into two equal parts so we can write,
BC = 2BD ……………(vii)
Now in equation (vi) we can substitute the result of (vii) so now we can write it as,
AC = 2BD
Which can be further written as,
$BD=\dfrac{AC}{2}$
Now squaring on both the sides we get,
$B{{D}^{2}}=\dfrac{A{{C}^{2}}}{4}$
Now putting $B{{D}^{2}}=\dfrac{A{{C}^{2}}}{4}$ in equation (v) we get,
$\begin{align}
& A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+B{{D}^{2}} \\
& \Rightarrow A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+\dfrac{A{{C}^{2}}}{4} \\
\end{align}$
Hence we can write,
$A{{B}^{2}}+D{{C}^{2}}=\dfrac{5}{4}A{{C}^{2}}$
Hence proved
Note: Students generally stuck after reading to the equation $A{{B}^{2}}+D{{C}^{2}}=A{{C}^{2}}+B{{D}^{2}}.$
They generally forget to use the property that the altitude, median and angle bisector are aligned in the same line. So they should be careful about it.
Last updated date: 28th Sep 2023
•
Total views: 362.7k
•
Views today: 8.62k
Recently Updated Pages
What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

How many millions make a billion class 6 maths CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

What is the past tense of read class 10 english CBSE

How many crores make 10 million class 7 maths CBSE

How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
