Q. 10

Two straight lines parallel to the sides BC and BA respectively through two vertices A and C of ΔABC meet at D. Let’s prove that ABC = ADC.

Answer :

Given: ΔABC, line l||BC, line m||BA, line l and line m are meeting at point D


To prove: ABC = ADC


The figure for the given question is as shown below,



Now in the given figure BA is parallel to line m with AC as tranversal line, so


BAC = ACD……….(i) (as they form is alternate interior angles)


Similarly, BC is parallel to line l with AC as tranversal line, so


ACB = CAD……….(ii) (as they form is alternate interior angles)


Now consider ΔABC,


We know in a triangle the sum of all three interior angles is equal to 180°.


So in this case,


ABC + BAC + ACB = 180°


Substituting the values from equation(i) and (ii), we get


ABC + ACD + CAD = 180°


ABC = 180° - ACD - CAD…………(iii)


Now consider ΔADC,


We know in a triangle the sum of all three interior angles is equal to 180°.


So in this case,


ADC + CAD + ACD = 180°


ADC = 180° - CAD - ACD…..(iv)


Equating equation (iii) and equation (iv), we get


ABC = ADC


Hence proved


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