Question

Please solve the following

- AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB
- l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that triangle ABC is congruent to triangle CDA

Answer 1

Answer:

1.)Given,

AD and BC are perpendiculars of AB.

TOBESHOWN :

CD bisects AB

∠BOC = ∠DOC ( ∴ Vertically opposite angles )

DA = BC

∠B = ∠A = 90°

So, by \bf{AAS}AAS congruence condition, ΔBOC ≅ ΔOAD

So,

now CO = OD

so, it bisects AB on point '\bf{O}O '

OA = OB [ by \bf{CPCT}CPCT ]

➖➖➖➖➖➖➖➖➖➖➖➖➖

2.) 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

Answer:

Given: l || m and p || q

To Prove: ΔABC ≅ ΔCDA.

We can show both the triangles are congruent by using ASA congruency criterion

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

In ΔABC and ΔCDA,

∠BAC = ∠DCA (Alternate interior angles, as p and q are parallel lines)

AC = CA (Common)

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

∴ ΔABC ≅ ΔCDA (By ASA congruence rule)