In an isosceles triangle ABC, with AB=AC, BD is the perpendicular bisector from B to side AC Prove that - Maths -

Question

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

Aakash Sharma · Accepted Answer

Given: ABC is an isosceles âˆ† with AB = AC = a (say) (1) and BD is the perpendicular bisector of AC Since perpendicular from B bisects AC or median BD is âŠ¥ to AC âˆ´ AB = BC (2) from (1) and (2) AB = BC = AC = a Hence ABC is an equilateral triangle Now we know that altitude of an equilateral âˆ†of side a is $\frac{\sqrt{3}}{2} a$ $\Rightarrow BD = \frac{\sqrt{3}}{2} a and CD = AD = \frac{1}{2} AC = \frac{a}{2}$ ${Thus BD}^{2} - {CD}^{2} = (\frac{\sqrt{3}}{2} a)^{2} - (\frac{a}{2})^{2} = \frac{3}{4} a^{2} - \frac{a^{2}}{4} = \frac{2}{4} a^{2} = 2 \times \frac{a}{2} \times \frac{a}{2} = 2 \times CD \times AD$

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

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