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Illustrative Examples Based on Congruence of Triangle

Example 1 :

In ΔABC, AB = AC = 8 cm and BAC = 110°, find B and C.

Solution

In ΔABC, since AB = AC 
... ∠B = C ...... (Angles opposite to equal sides)


 

Now, A + B + C = 180o ...... (Sum of three angles of a triangle)
= 110° + B + B = 180°
= 2 B = 180° - 110°

                            = 70°
= B = 35°
...B = C = 35°


 

 

 
Example 2 :

In the figure, AB = AC and MB = MC. Prove that ABM = ACM

Solution

In ΔABC, AB = AC (given)
...       ABC = ACB ...... (i) (Angles opposite to equal sides)
Now in
ΔMBC, MB = MC
...       MBC = MCB ...... (ii)
Subtracting eqn (ii) from eqn (i), we get
     
ABC - MBC = ACB - MCB
...              ABM = ACM

 

Example 3 :

Prove that ΔABC is isosceles if
(i) Altitude AD bisects BC
(ii) Median AD is perpendicular to BC

Solution
 

(i) Since AD BC, ADB = ADC = 90°
Now in
ΔABD and ΔACD, we have
ADB = ADC ... (each 90°)
AD = AD ... (common)
BD = DC ... (given)

ΔADB ΔADC
AB = AC 

ΔABC is an isosceles triangle

(ii) Solve as in (i)

(As ADB = ADC =180°/2 = 90°)





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