TRIANGLES
LONG ANSWER QUESTIONS
1. A point O is taken inside an equilateral four sided figure ABCD such
that its distance from the angular points D and B are equal . show that
AO and OC are in one and the same straight line .
Sol:
Given,
A point O inside an equilateral four sided figure ABCD such that
BO = OD
To prove
AO and OC are in one end and the same straight line .
Proof:
In △s AOD and AOB
AD = AB (given)
AO=AO (common side )
OD = OB ( given)
△ AOD ≅ △AOB ( by SSS congruency)
⇒ ∠1= ∠2 ……….(1) (C.P.C.T)
C.P.C.T ⇒ CORRESPONDING PARTS OF CONCURRENT TRIANGLES
Similarly , ∠3 = ∠4……….2
But ∠1+∠2+∠3+∠4 = 4 right angles .
Because
Sum of the angles at a point it is 4 right angles .
By using 1 and 2
∴ 2∠2 + 2∠3 = 4 𝑟𝑖𝑔ℎ𝑡 𝑎𝑛𝑔𝑙𝑒𝑠
2(∠2+∠3) = 4 right angles
(∠2+∠3) = 2 right angles
(∠2+∠3) = 1800
∠2 and ∠3 forms a linear pair
∴ AO and OC are in the same straight line so , AC is a straight line .
SHORT ANSWER QUESTIONS
1. In figure AD = BC AND BD = CA . Prove that
∠ADB = ∠ BCA and ∠DAB = ∠CBA
SOL: In triangles ABD and BAC we have
AD = BC (given)
BD = AC (given)
AB = BA (given)
∴ △ABD ≅ △BAC ( by SSS congruency)
⇒ ∠ DAB = ∠ CBA
And ∠ ADB = ∠BCA ( by C.P.C.T)
2. In a quadrilateral ABCD , AB = AD and BC = CD show that ∠ABC =
∠ADC.
SOL: First join BD.
Now in △ABD
AB = AD(given)
∠1 = ∠2 ……..(1)
(angles opposite to equal sides are equal)
Similarly, ∠3= ∠4 ……(2)
By adding (1) and (2)
∠1 + ∠3 = ∠2 + ∠4
⇒ ∠ABC = ∠ADC.