Question

l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that ΔABC ≅ ΔCDA.

Can we answer in the following way? If not, Please guide me.
In ΔABC and ΔCDA,

AC=CA {common}
AB=CD {Since two set of parallel lines are intersected by another set of parallel lines}
BC=AD {Since two set of parallel lines are intersected by another set of parallel lines}
So ΔABC ≅ ΔCDA, (SSS Congruence)

Answer 1

Nope Bro!!

Let it be like this

This one's easy

Since l and m are 2 parallel lines and AC is the transversal

Therefore, AD || BC and AC is the transversal

Therefore, /_DAC = /_ACA (Alternate Angles)

Also /_BAC = /_ACD (Alternate Angles)

Now, In ∆ABC & ∆ CDA

/_DAC = /_ACA (Alternate Angles)

/_BAC = /_ ACD (Alternate Angles)

AC = CA (Common)

Therefore, ∆ABC ≅ ∆CDA (By ASA Congruence rule)

Answer 2

Step-by-step explanation:

$\huge\color{Red}{\colorbox{black}{XxItzAdarshxX࿐ ❥︎}} </p><p>$

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, triangle ABC triangle CDA.

$\huge\large\bf{\underline\green{❥thαnk \; чσu ♥♥}}$