Question

In an isosceles triangle ABC with AB=AC, BD is perpendicular from B to the side AC. Prove that BD2-CD2=2AD.CD

Answer 1

▶ Question :-

→ In an isosceles ∆ABC with AB = AC, BD is perpendicular from B to the side AC. Prove that BD² - CD² = 2AD.CD .


$\huge \pink{ \mid{ \underline{ \overline{ \sf Solution :- }} \mid}}$


▶ Given :-

→ A ∆ABC in which AB = BC and $BD \perp AC$ .


▶ To prove :-

→ ( BD² - CD² ) = 2AD.CD .


▶ Proof :—


From right ∆ADB, we have

→ AB² = AD² + BD² . [ By Pythagoras theorem ] .


==> AC² = AD² + BD² [ °•° AB = AC ] .

==> ( AD + CD )² = AD² + BD² [ °•° AC = CD + AD ] .

==> AD² + CD² + 2AD.CD = AD² + BD² .

==> 2AD.CD = AD² + BD² - AD² - CD² .

$\boxed{ \green{ \sf \therefore ( BD^2 - CD^2 ) = 2AD.CD . }}$



✔✔ Hence, it is proved ✅✅.



$\huge \boxed{ \blue{ \mathbb{THANKS}}}$

Answer 2

From right ∆ADB, we have

→ AB² = AD² + BD² . [ By Pythagoras theorem ] .


AC² = AD² + BD² .

( AD + CD )² = AD² + BD² .

AD² + CD² + 2AD.CD = AD² + BD² .

2AD.CD = AD² + BD² - AD² - CD² .

→ BD² - CD² = 2AD.CD .