Question
In $\triangle\text{ABC},$ given that $AB = AC$ and $\text{BD}\perp\text{AC}.$
Prove that $BC^2 = 2AC \times CD.$
Prove that $BC^2 = 2AC \times CD.$
✓
Answer
Given: In $\triangle\text{ABC},$ $\text{AB} = \text{AC, } \text{BD}\perp\text{AC}$
To prove: $BC^2 = 2AC \times CD$
$\therefore$ $AB^2 = BD^2 + AD^2$ (Pythagoras Theorem)
$\Rightarrow BD^2 = AB^2 - AD^2 .....(i)$
Similarly in right $\triangle\text{BDC},$
$BC^2 = BD^2 + DC^{2$
$} = AB^2 - AD^2 + DC^2 [From (i)] $
$= AB^2 - (AC - CD)^2 + CD^2 $
$= AB^2 - (AC^2 + CD^2 - 2AC \times CD) + CD^2 $
$= AC^2 - AC^2 - CD^2 + 2AC \times CD + CD^2$
($\because$ $AB = AC$) $= 2AC × CD$ (given)
Hence $BC^2 = 2AC \times CD$
To prove: $BC^2 = 2AC \times CD$
$\therefore$ $AB^2 = BD^2 + AD^2$ (Pythagoras Theorem)
$\Rightarrow BD^2 = AB^2 - AD^2 .....(i)$
Similarly in right $\triangle\text{BDC},$
$BC^2 = BD^2 + DC^{2$
$} = AB^2 - AD^2 + DC^2 [From (i)] $
$= AB^2 - (AC - CD)^2 + CD^2 $
$= AB^2 - (AC^2 + CD^2 - 2AC \times CD) + CD^2 $
$= AC^2 - AC^2 - CD^2 + 2AC \times CD + CD^2$
($\because$ $AB = AC$) $= 2AC × CD$ (given)
Hence $BC^2 = 2AC \times CD$
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
Four equal circles, each of radius a, touch each other. Show that the area between them is $\frac{6}{7}\text{a}^2$ $\Big(\text{Take }\pi=\frac{22}{7}\Big).$
→Solve for x and y:
$\text{x}+\frac{5}{\text{y}}=\text{6},$
$\text{3x}-\frac{8}{\text{y}}=\text{5}\ (\text{y}\neq0).$
→$\text{x}+\frac{5}{\text{y}}=\text{6},$
$\text{3x}-\frac{8}{\text{y}}=\text{5}\ (\text{y}\neq0).$
Find the sum of all natural numbers between $1$ and $100$, which are divisible by $3$.
→How many numbers lie between $10$ and $300$, which when divided by $4$ leave a remainder 3?
→The sum of the first n terms of an AP is $\Big(\frac{5\text{n}^2}{2}+\frac{3\text{n}}{2}\Big).$ Find the $n^{th}$ term and the $20^{th}$ term of this AP.
→Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.
Prove the following identities:
$\frac{\cos\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\cos\theta-\sin\theta)}=(\cos\theta+\sin\theta)$
→$\frac{\cos\theta}{(1-\tan\theta)}+\frac{\sin^2\theta}{(\cos\theta-\sin\theta)}=(\cos\theta+\sin\theta)$
Check whether 301 is in the sequence
$5,11,17,23, \ldots \text { ? }$
→$5,11,17,23, \ldots \text { ? }$
Solve : m² - 14 m + 13 = 0
Prove the following trigonometric identities.
If $\text{a }\cos^3\theta+3\text{a}\cos\theta\sin^2\theta=\text{m, a }\sin^3\theta+3\text{a}\cos^2\theta\sin\theta=\text{n},$ prove that $(\text{m}+\text{n})^\frac{2}{3}+(\text{m}-\text{n})^\frac{2}{3}=2\text{a}^\frac{2}{3}.$
→If $\text{a }\cos^3\theta+3\text{a}\cos\theta\sin^2\theta=\text{m, a }\sin^3\theta+3\text{a}\cos^2\theta\sin\theta=\text{n},$ prove that $(\text{m}+\text{n})^\frac{2}{3}+(\text{m}-\text{n})^\frac{2}{3}=2\text{a}^\frac{2}{3}.$