In the given fig., D is a point on hypotenuse AC of ∆ABC, DIM ⊥ BC and DN ⊥ AB.
Prove that:
(i) DM2 = DN × MC.
(ii) DN2 = DM × AN.
We have,
AB ⊥ BC and DM ⊥ BC
⇒ AB || DM
Similarly, we have
CB ⊥ AB and DN ⊥ AB
⇒ CB || DN
Hence, quadrilateral BMDN is a rectangle.
∴ BM = ND
(i) In ∆BMD, we have
∠1 + ∠BMD + ∠2 = 180°
⇒ ∠1 + 90° + ∠2 = 180°
⇒ ∠1 + ∠2 = 90°
Similarly, in ∆DMC, we have
∠3 + ∠4 = 90°
Since BD ⊥ AC. Therefore
∠2 + ∠3 = 90°
Now, ∠1 + ∠2 = 90°and ∠2 + ∠3 = 90°
⇒ ∠1 + ∠2 = ∠2 + ∠3
⇒ ∠1 = ∠3
Also, ∠3 + ∠4 = 90° and ∠2 + ∠3 = 90°
⇒ ∠3 + ∠4 = ∠2 + ∠3 ⇒ ∠2 = ∠4
Thus, in ∆'s BMD and DMC, we have
∠1 = ∠3 and ∠2 = ∠4
Therefore, by using AA similar condition
(ii) Proceeding as in (i), we can prove that
In the given fig., ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2 BC . BD.
Given: An obtuse triangle ABC, obtuse angled at B, and AD ⊥ CB produced.
To Prove : AC2 = AB2 + BC2 + 2BC.BD
Proof: In right triangle ADB, we have
AB2 = AD2 + BD2 ...(i)
[Using Pythagoras theorem]
In ∆ADC, we have
AC2 = AD2 + DC2
⇒ AC2 = AD2 + (DB + BC)2
⇒ AC2 = AD2 + DB2 + BC2 + 2DB.BC
⇒ AC2 = AB2 + BC2 + 2BC.BD
[from (i), we have AB2 = AD2 + BD2]
In the given Fig, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC. Prove that AC2= AB2 + BC2 - 2 BC.BD.
Given: An acute triangle, acute angled at B and AD ⊥ BC.
To Prove : AC2 = AB2 + BC2 — 2BC.BD
Proof: In ∆ABD, we have
AB2 = AD2 + BD2 ...(i)
[Using Pythagoras theorem]
In ∆ADC, we have
AC2 = AD2 + DC2
⇒ AC2 = AD2 + (BC - BD)2
⇒ AC2 = AD2 + BC2 - 2BC.BD + BD2
⇒ AC2 = (AD2 + BD2) + BC2 - 2BC.BD
⇒ AC2 = AB2 + BC2 - 2BC.BD
[from (i), we have AD2 + BD2 = AB2]
In the given fig, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i)
(ii)
(iii)
Given: ABCD is a parallelogram in which AC and BD are diagonals which intersect each other at O.
To Prove: AB2 + BC2 + CD2 + DA2 = AC2 + BD2
Proof: Since diagonals of parallelogram bisect each other.
Therefore, BO and DO are medians of ∆ABC and ∆ADC, respectively.
We know that the sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.
Now, in ∆ABC,
∵ OB is a median