Mathematics
In an isosceles triangle ABC; AB = AC and D is a point on BC produced. Prove that :
AD2 = AC2 + BD.CD
Pythagoras Theorem
12 Likes
Answer
By formula,
By pythagoras theorem,
⇒ (Hypotenuse)2 = (Perpendicular)2 + Base2
Draw AE perpendicular to BC.
In right angle triangle AED,
By pythagoras theorem,
⇒ AD2 = AE2 + ED2
⇒ AD2 = AE2 + (EC + CD)2 ………(1)
In right angle triangle AEC,
By pythagoras theorem,
⇒ AC2 = AE2 + EC2
⇒ AE2 = AC2 - EC2 ………(2)
Substituting value of AE2 from equation (2) in (1), we get :
⇒ AD2 = AC2 - EC2 + (EC + CD)2
⇒ AD2 = AC2 - EC2 + EC2 + CD2 + 2.EC.CD
⇒ AD2 = AC2 + CD(CD + 2EC) ……….(3)
Since, ABC is an isosceles triangle.
We know that,
In an isosceles triangle altitude from the vertex bisects the base.
∴ E is the mid-point of BC.
⇒ EC = BC
⇒ BC = 2EC.
From figure,
⇒ BD = BC + CD
⇒ BD = 2EC + CD
Substituting above value in equation (3), we get :
⇒ AD2 = AC2 + CD.BD
Hence, proved that AD2 = AC2 + BD.CD
Answered By
7 Likes
Related Questions
Diagonals of rhombus ABCD intersect each other at point O. Prove that :
OA2 + OC2 = 2AD2 -
In the figure, AB = BC and AD is perpendicular to CD. Prove that :
AC2 = 2.BC.DC.
In triangle ABC, angle A = 90°, CA = AB and D is a point on AB produced. Prove that :
DC2 - BD2 = 2AB.AD.
In triangle ABC, AB = AC and BD is perpendicular to AC. Prove that :
BD2 - CD2 = 2CD × AD