Question

Question: ABC is an isosceles triangle in which AB=AC. AD bisects exterior angle PAC and CD is parallel to AB. Show that angle DAC = angle BCA. and ABCD is a parallelogram.

Guys please tell if I have done it correct?

Answer 1

Answer:

yes mate, it's correct!

Step-by-step explanation:

Answer 2

Given:

ABC is an isosceles triangle.

AB = AC

To Find:

ABCD is a parallelogram.

Solution:

AD bisects ∠PAC

So, ∠PAD = ∠DAC = 1/2∠PAC

In ΔABC,

AB = AC [given]

∠ABC = ∠BCA [angles opposite to equal sides of a triangle are equal] ..(ii)

∠PAC= ∠ACB+∠BCA [exterior angle is the sum of interior opposite

angles]

An exterior angle of a triangle is equal to the two interior opposite angles.

⇒ ∠PAC = ∠BCA + ∠BCA [using(ii)]

⇒ 2∠CAD = 2∠BCA [∵AB bisects PAC]

⇒ ∠CAD = ∠BCA

⇒ ∠DAC = ∠BCA [using(i) ∠DAC=1/2∠PAC]

These angles are from a pair of equal alternate interior angles.

∴ AD║BC

CD║AB

Therefore, ABCD is a parallelogram.