Question
In an equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$ prove that $AD^2 = 3BD^2$.
✓
Answer
Given: $\triangle\text{ABC}$ is an equilateral in which $AB = BC = CA.$
$\text{AD}\perp\text{BC}$
To prove: $AD^2 = 3BD^2$
Proof: In equilateral $\triangle\text{ABC}.$
$\because\text{AD}\perp\text{BC}$
$\therefore$ AD bisects $BC$ at $D$
$\therefore\text{BD}=\frac{1}{2}\text{BC}$
Now in right $\triangle\text{ABD}$
$AB^2 = AD^2 + BD^2$ (pythagoras theorem)
$\therefore$ $AD^2 = AB^2 - BD^2$
$AD^2 = BC^2 - BD^2$ ${AB = AC = BC$ given$}$
$AD^2 = (2BD)^2 - BD^2 {BC = 2BD}$
$AD^2 = 4BD^2 - BD^2$
$AD^2 = 3BD^2$
Hence proved.
$\text{AD}\perp\text{BC}$
To prove: $AD^2 = 3BD^2$
Proof: In equilateral $\triangle\text{ABC}.$
$\because\text{AD}\perp\text{BC}$
$\therefore$ AD bisects $BC$ at $D$
$\therefore\text{BD}=\frac{1}{2}\text{BC}$
Now in right $\triangle\text{ABD}$
$AB^2 = AD^2 + BD^2$ (pythagoras theorem)
$\therefore$ $AD^2 = AB^2 - BD^2$
$AD^2 = BC^2 - BD^2$ ${AB = AC = BC$ given$}$
$AD^2 = (2BD)^2 - BD^2 {BC = 2BD}$
$AD^2 = 4BD^2 - BD^2$
$AD^2 = 3BD^2$
Hence proved.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solve the following quadratic equations by factorization:
$25x(x + 1) = -4$
→$25x(x + 1) = -4$
Solve the following simultaneous equations.
$5 x-3 y=8 ; 3 x+y=2$
→$5 x-3 y=8 ; 3 x+y=2$
Find the roots of the following equations, if they exist, by applying the quadratic formula:
$x^2 - 6x + 4 = 0$
→$x^2 - 6x + 4 = 0$
Solve the following simultaneous equation graphically.
2x – 3y = 4 ; 3y – x = 4
2x – 3y = 4 ; 3y – x = 4
Draw a circle of radius 3 cm and draw chord XY 5 cm long. Draw the tangent of the circle passing through point X and point Y (without using the center of the circle)
A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is:
- Red.
- Black.
If m times the mth term of an A.P. is equal to n times nth term then show that the (m + n)th term of the A.P. is zero.
A tree was broken due to storm. Its broken upper part was so inclined that its top touched the ground making an angle of 30°with the ground.The distance from the foot of the tree and the point where the top touched the ground was 10 metre. What was the height of the tree.
12 defective pens are accidently mixed with 132 good ones. It is not possible to just look at pen and tell whether or not it is defective. one pen is taken out at random from this lot. Determine the probability that the pen taken out is good one.
A horse is placed for grazing inside a reangular field 70m by 2m. It is tethered to one comer by a rope 21m long. On how much area an it graze? How much area is left ungrazed? $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→