Question

in a right angled triangle ABC, angle B=90degree, BD is perpendicular to AC, and AD:CD=3:2. then prove that 2BC^2=5CD^2

Answer 1

To prove $2BC^{2}=5CD^{2}$in a right triangle ABC and BD⊥AC;AD:CD=3:2

Step-by-step explanation:

• Given:-

1. ABC is a right angled triangle
2. ∠B=90°
3. BD⊥AC
4. AD:CD=3:2

• To prove:-

$2BC^{2} =5CD^{2}$

• Proof:-

Pythagoras theorem-$hypotenuse^{2} =perpendicular^{2} +base^{2}$

$AB^{2} +BC^{2} =AC^{2}$

Since, $AB^{2} =BD^{2} +AD^{2}$ and $AC^{2} =AD^{2} +CD^{2}$

So,($(BD^{2} +AD^{2} )+BC^{2} =(AD+CD)^{2}$

=>$Since,[AD=\frac{3}{2}]\[BD^{2} +(\frac{3}{2})^{2}]+BC^{2} =[\frac{3}{2}CD+CD]^{2}$

=>$BC^{2}-CD^{2}+\frac{9}{4}CD^{2}+BC^{2}=\frac{25}{4}CD^{2}$

=>$2BC^{2}=5CD^{2}$

Hence proved.