MATH SOLVE

4 months ago

Q:
# Let ABCD be a cyclic quadrilateral. Let P be the intersection of AD and BC, and let Q be the intersection of AB and CD. Prove that the angle bisectors of angle DPC and angle AQD are perpendicular.

Accepted Solution

A:

Given:

Cyclic quadrilateral ABCD.

Produce AD and BC to meet at P.

Produce AB and CD to meet at Q,

Angle bisector PJ of angle DPC divides angle DPC into two equal parts,x.

Angle bisector QI of angle AQD divides angle AQD into two equal parts,y.

Solution:

Construct segment PQ.

Name U= angle BPQ

Name V=angle BQP.

See attached figure.

Steps:

1. Consider triangle BPQ:

U+V=angle ABP (exterior angle)........................(1)

2. Consider triangle DPQ

U+2X+V+2Y = 180-angle PDQ (exterior angle)

= angle ABC (cyclic quad->opposite angles are supplementary)

=>

U+2X+V+2Y = angle ABC...................................(2)

3. Add (1) and (2)

U+V + U+2X+V+2Y = angle DPQ+angle ABC

=>

2(U+V+X+Y)=180 degrees (angles ABP and ABC lie on a straight lie)

=>

U+V+X+Y = 90 degrees = angle IKJ

4. angle bisectors PJ and QI are perpendicular.

Cyclic quadrilateral ABCD.

Produce AD and BC to meet at P.

Produce AB and CD to meet at Q,

Angle bisector PJ of angle DPC divides angle DPC into two equal parts,x.

Angle bisector QI of angle AQD divides angle AQD into two equal parts,y.

Solution:

Construct segment PQ.

Name U= angle BPQ

Name V=angle BQP.

See attached figure.

Steps:

1. Consider triangle BPQ:

U+V=angle ABP (exterior angle)........................(1)

2. Consider triangle DPQ

U+2X+V+2Y = 180-angle PDQ (exterior angle)

= angle ABC (cyclic quad->opposite angles are supplementary)

=>

U+2X+V+2Y = angle ABC...................................(2)

3. Add (1) and (2)

U+V + U+2X+V+2Y = angle DPQ+angle ABC

=>

2(U+V+X+Y)=180 degrees (angles ABP and ABC lie on a straight lie)

=>

U+V+X+Y = 90 degrees = angle IKJ

4. angle bisectors PJ and QI are perpendicular.