Question

In ΔABC, AD bisects the exterior angle ∟PAC at A and AD is parallel to BC. Prove that ΔABC is isosceles.​

Answer 1

Answer:

(i)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCA

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2y

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)(ii)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)(ii)Now because,

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)(ii)Now because,∠DAC=∠BCA (proved above)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)(ii)Now because,∠DAC=∠BCA (proved above)Therefore , AD∣∣BC

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)(ii)Now because,∠DAC=∠BCA (proved above)Therefore , AD∣∣BCAnd CD∣∣BA (Given)

(i)∠ABC=∠BCA=y(let) (Because triangle ABC is an isosceles triangle)∠QAD=∠DAC=x(let) (Given)∠DCA=∠BAC=z(let) (Alternate interior angles)And we know that an exterior angle of a triangle is equal to the sum of the opposite interior angles.So,∠QAD+∠DAC=∠ABC+∠BCAx+x=y+y2x=2yx=y∠DAC=∠BCA (hence proved)(ii)Now because,∠DAC=∠BCA (proved above)Therefore , AD∣∣BCAnd CD∣∣BA (Given)Since opposite sides of quadrilateral ABCD are parallel therefore ABCD is a parallelogram.

Answer 2

Answer:In ΔABC, AD bisects the exterior angle ∟PAC at A and AD is parallel to BC. Prove that ΔABC is isosceles.​

Step-by-step explanation: