Question

if ABC is an isosceles triangle and D is a point on BC such that AD is perpendicular to BC ,then proove that AB square - AD square equals to BD.DC​

Answer 1

Step-by-step explanation:

see the pic...........................

Answer 2

$AB^2-AD^2=BD\cdot CD$          

Hence prove

Step-by-step explanation:

If ABC is an isosceles triangle and D is a point on BC

Such that AD is perpendicular to BC

D is mid point of BC

$AD=\dfrac{1}{2}BC$

Please find attachment for figure.

In ΔABD, ∠ADB = 90°

$AB^2=AD^2+BD^2$        By using Pythagoras theorem

$AB^2-AD^2=\dfrac{1}{4}BC^2$                             $\because AD=\dfrac{1}{2}BC$

$AB^2-AD^2=\dfrac{1}{4}(BD+DC)^2$               $\because BC=BD+DC$

$4AB^2-4AD^2=BD^2+DC^2+2BD\cdot CD$     $\because (a+b)^2=a^2+b^2+2ab$

$4AB^2-4AD^2=AB^2-AD^2+AC^2-AD^2+2BD\cdot CD$

$2AB^2-2AD^2=2BD\cdot CD$                   $\because AB=AC$

$AB^2-AD^2=BD\cdot CD$          

Hence prove

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