Question
In the given figure, if $AD \bot BC$, prove that $AB^2 + CD^2 = BD^2 + AC^2.$
✓
Answer
In right angled $\triangle BDA$,
By pythagoras theorem
$AB^2 = AD^2 + BD^2 ...(i)$
And in right angled $\triangle CDA$ ,
By pythagoras theorem
$AC^2 = CD^2 + AD^2 ...(ii)$
On subtracting Eq(ii) from Eq(i) , we get
$AB^2 - AC^2 = [AD^2 + BD^2] - [CD^2 + AD^2]$
$AB^2 - AC^2 = AD^2 + BD^2 - CD^2 - AD^2$
$AB^2 - AC^2 = BD^2 - CD^2$
$\therefore $ $AB^2 + CD^2 = BD^2 + AC^2$
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
Two vertices of a triangle have co-ordinates (-8, 7) and (9, 4). If the centroid of the triangle is at the origin, what are the co-ordinates of the third vertex?
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
A solid sphere of radius $10.5 \ cm$ is melted and recast into smaller solid cones, each of radius $3.5 \ cm$ and height $3 \ cm .$ find the number of cones so formed. $($Use $\pi=\frac{22}{7})$
→Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.
There are 35 students in a class of whom 20 are boys and 15 are girls. From these students one is chosen at random. What is the probability that the chosen student is:
-
Boy.
-
Girl.
Solve the following quadratic equation:$\sqrt3\text{x}^2+\text{11x}+6\sqrt3=0$
→The first and the last terms of an $AP$ are $5$ and $45$ respectively. If the sum of all its terms is $400 ,$ find its common difference.
→State whether the following quadratic equations have two distinct real roots. Justify your answer.
$3x^2 - 4x + 1 = 0.$
→$3x^2 - 4x + 1 = 0.$
The circumforence of a circle is 39.6cm. Find its area. $\Big[\text{Use }\pi=\frac{22}{7}\Big]$
→In Figure $,AB$ is diameter of a circle centered at $O. B C$ is tangent to the circle at $B$. If $O P$ bisects the chord $A D$ and $\angle A O P=60^{\circ}$, then find $m \angle C$.
→