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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