ABCD is a parallelogram.the bisectors of the angle and C meet the diagonal BD in P and Q respectively. Prove that triangleAPQ ~=triangleCOD.
To prove that triangle APQ is similar to triangle COD, we need to show that their corresponding angles are congruent and their corresponding sides are proportional.
First, let's consider the angle congruence:
In parallelogram ABCD, opposite angles are congruent. Therefore, angle A = angle C and angle B = angle D.
Let α be the measure of angle A (which is also the measure of angle C since they are congruent). Since the bisector of angle C intersects diagonal BD at point Q, we know that angle PQD is half of angle C.
Therefore, angle PQD = α/2.
Now, let's consider triangle COD. Angle ODC is a straight angle (180 degrees), so angle DOC = 180 - α.
Since P is on the diagonal BD, we can say angle BPD = angle DPB = 180 - α.
Since angle PQD is half of angle C, we have angle DPQ = α/2.
Now, let's consider the side proportionality:
In triangle APQ, we know that angle PQD = α/2. We also know that angle DPQ = α/2.
Therefore, triangle APQ is an isosceles triangle, and AP = AQ.
Now, let's consider triangle COD. Angle DOC = 180 - α. We also know that angle BPD = angle DPB = 180 - α.
Therefore, triangle COD is an isosceles triangle, and CD = OD.
Since AP = AQ and CD = OD, we can conclude that triangle APQ is similar to triangle COD by the Side-Side-Side (SSS) similarity criterion.
Therefore, we have proven that triangle APQ is similar to triangle COD.
To prove that triangle APQ is similar to triangle COD, we need to show that their corresponding angles are equal.
Since ABCD is a parallelogram, the opposite angles are equal. Therefore, angle DAB is equal to angle BCD.
Let's consider the angle bisectors of angle A and angle C. The angle bisector of angle A meets the diagonal BD at point P, and the angle bisector of angle C meets the diagonal BD at point Q.
Since P is the intersection of the angle bisector of angle A and the diagonal BD, angle APB is equal to angle DAB/2.
Similarly, since Q is the intersection of the angle bisector of angle C and the diagonal BD, angle BQC is equal to angle BCD/2.
Now let's consider triangles APQ and COD.
In triangle APQ:
- Angle PAQ is equal to angle DAB/2 (as discussed earlier)
- Angle PQA is equal to angle ABQ (opposite angles in parallelogram ABCD)
In triangle COD:
- Angle POC is equal to angle BCD/2 (as discussed earlier)
- Angle CDO is equal to angle BQC (opposite angles in parallelogram ABCD)
Therefore, we can conclude that angle PAQ is equal to angle POC, and angle PQA is equal to angle CDO. By the Angle-Angle Similarity Postulate, we can then conclude that triangle APQ is similar to triangle COD.