Class 9 SELINA Solutions Maths Chapter 9 - Triangles [Congruency in Triangles]
Triangles [Congruency in Triangles] Exercise Ex. 9(A)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
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Solution 12
Solution 13
Solution 14
Solution 15
Triangles [Congruency in Triangles] Exercise Ex. 9(B)
Solution 1
Solution 2
Solution 3
Solution 4
Solution 5
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Solution 12
Solution 13
Solution 14
Solution 15
In triangles AOE and COD,
∠A = ∠C (given)
∠AOE = ∠COD (vertically opposite angles)
∴ ∠A + ∠AOE = ∠C + ∠COD
⇒180° - ∠AEO = 180° - ∠CDO
⇒ ∠AEO = ∠ CDO ….(i)
Now, ∠AEO + ∠OEB = 180° (linear pair)
And, ∠CDO + ∠ODB = 180° (linear pair)
∴ ∠AEO + ∠OEB = ∠CDO + ∠ODB
⇒ ∠OEB = ∠ODB [Using (i)]
⇒ ∠CEB = ∠ADB ….(ii)
Now, in ΔABD and ΔCBE,
∠A = ∠C (given)
∠ADB = ∠CEB [From (ii)]
AB = BC (given)
⇒ ΔABD ≅ ΔCBE (by AAS congruence criterion)
Solution 16
In ΔAOD and ΔBOC,
∠ AOD = ∠ BOC (vertically opposite angles)
∠ DAO = ∠ CBO (each 90°)
AD = BC (given)
∴ ΔAOD ≅ ΔBOC (by AAS congruence criterion)
⇒ AO = BO [cpct]
⇒ O is the mid-point of AB.
Hence, CD bisects AB.
Solution 17
In ΔABC,
AB = AC
⇒ ∠B = ∠C (angles opposite to equal sides are equal)
Solution 18
Solution 19
Solution 20
In ΔABD and ΔBAC,
AD = BC (given)
BD = CA (given)
AB = AB (common)
∴ ΔABD ≅ ΔBAC (by SSS congruence criterion)
⇒ ∠ADB = ∠BCA and ∠DAB = ∠CBA (cpct)
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