Q6 of 27 Page 289

ABC is an equilateral triangle. AD is perpendicular on the side BC, let us prove that, AB2 + BC2 + CA2 = 4AD2.


Given: AB = BC = CA = x


∠ADC = ∠ADB = 90°


To prove: AB2 + BC2 + CA2 = 4AD2


Proof:


We know that in an equilateral triangle perpendicular from any vertex on opposite side bisects it.


So, BD = DC = 1/2BC ……………… (1)


By applying Pythagoras theorem in ΔABD, we get,


AB2 = AD2 + BD2 [∵ H2 = P2 + B2]


By substituting BD from eqn. (1) we get,


AB2 = AD2 + (1/2BC)2


⇒ AB2 = AD2 + 1/4BC2


⇒ 4AB2 = 4AD2 + BC2


⇒ 4AB2 - BC2 = 4AD2


⇒ 4AB2 - AB2 = 4AD2 [Given: AB = BC]


⇒ 3AB2 = 4AD2


⇒ AB2 + BC2 + CA2 = 4AD2 [∵ AB = BC = CA]


Hence proved.


More from this chapter

All 27 →
4

I have drawn a triangle PQR whose ∠Q is right angle. If S is any point on QR, let us prove that, PS2 + QR2 = PR2 + QS2

 5

Let us prove that, the sum of squares drawn on the sides of a rhombus is equal to the sum of squares drawn on two diagonals.

 7

I have drawn a right angled triangle ABC whose ∠A is right angle. I took two points P and Q on the sides AB and AC respectively. By joining P,B,Q and C, P let us prove that, BQ2 + PC2 = BC2 + PQ2

 8

If two diagonals of a quadrilateral ABCD intersect each other perpendicularly, then let us prove that, AB2 + CD2 = BC2 + DA2