Question

prove that the sides of triangle ABC are equal to the sides of triangle ACD​

Answer 1

Answer:

it is a rectangle

Step-by-step explanation:

AB=DC (opposite sides are equal)

AD=BC (opposite sides are equal)

H=x

P=2

B=6

h^2=p^2+b^2

h^2=6^2 + 2^2

h^2=36+4

h^2=40

h=√40

h=2√10

AC=2√10

Answer 2

Explanation:

We can solve this problem by several congruence conditions, but I'll go over a few of them.

ABCD is a rectangle. Hence, it only natural that CD = AB = 2cm and AD = BC = 6cm, for opposite sides of a rectangle are always equal.

Hence,

AB = CD (proven)

BC = AD (proven)

AC = CA (common side)

Hence, we have directly proven that all the sides of the triangles are equal by proving ΔABC ≅ ΔCDA through Side-Side-Side congruence condition. Let's go over Angle-Side-Angle condition now.

∠ACB = ∠CAD [∵ AD ║BC, AC is the transversal intersecting the two lines ;alternate interior angles are equal]

Similarly, ∠BAC = ∠DCA [∵AB ║CD, AC transversal]

Hence,

∠ACB = ∠CAD (proven)

AC = CA (common side)

∠BAC = ∠DCA (proven)

Hence, ΔABC ≅ ΔCDA by ASA congruence condition.

∴ AB = CD, AC = CA and BC = DA (common parts of congruent triangles)

Hence, proven.

I hope this answer helps. :D