Question

I and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ≅ ΔCDA.

Answer 1

l || m Therefore, <DAC = <ACB And <DCA = <CAB.....(1)(alt int. Angles)  Now in ∆ABC AND ∆CDA, We have, AC=AC.          (Common) <DAC = <ACB <DCA = <CAB(from 1)  Therefore ΔABC ≅ ΔCDA. ( By ASA criteria) . Thanks

Answer 2

Solution:

It is given that p q and l m

To prove:

Triangles ABC and CDA are similar i.e. ΔABC ΔCDA

Proof:

Consider the ΔABC and ΔCDA,

(i) BCA = DAC and BAC = DCA Since they are alternate interior angles

(ii) AC = CA as it is the common arm

So, by ASA congruency criterion, ΔABC ΔCDA.