Question

Fi.7.14
4. I and are two parallel lines intersected by
another pair of parallel lines p and a
(see Fig. 7.19). Show that A ABC ACDA
Fix 7.19
5. Limel is the bisector of an angle Z A and B is my
point on L. BP and BQ are perpendiculars from B
to the arms of A (see Fig. 7.30). Showth
GA APB sa AQв
BP BQ or B is equidistant from the arms
A LA​

Answer 1

Answer:

In △ABC and △CDA

∠BAC=∠DCA (Alternate interior angles, as p∥q)

AC=CA (Common)

∠BCA=∠DAC (Alternate interior angles, as l∥m)

∴△ABC≅△CDA (By ASA congruence rule)

Answer 2

Step-by-step explanation:

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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.

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