Question

Через точку пересечения диагоналей параллелограмма ABCD проведена прямая пересекающая стороны AD и BC в точках P иQ. Докажите что AP=CQ

Answer 1

1 шаг. Пусть О - центр п-мма. тогда О∈PQ. Докажем, что PO = OQ.
Рассмотрим ΔAOQ и COP. Очевидно, диагональ АС проходит через О. Тогда 
∠QAO=∠PCO(т.к. BC||AD и AC-секущая)
∠AOQ=∠COP(вертикальные)
AO=OC(т.к. О-центр п-мма, а AC-диагональ)
--------------------------------------
ΔAOQ=ΔCOP
2 шаг. Рассмотрим 4-уг. APCQ: диагонали пересекаются в точке О и
СО=ОА
PO=OQ(из равенства ΔAOQ и COP)
Значит APCQ - параллелограмм (т.к. диагонали точкой пересечения делятся пополам). Значит AP=QC(как противоположные стороны параллелограмма)

Answer 2

Proof that AP = CQ in a Parallelogram

To prove that AP = CQ in a parallelogram ABCD, where a line intersects sides AD and BC at points P and Q respectively, we can use the properties of parallelograms and the fact that the diagonals of a parallelogram bisect each other.

Step 1: Recall that in a parallelogram, opposite sides are parallel and congruent. Therefore, we have AB || CD and AD || BC.

Step 2: Since AB || CD, we can use the property of transversals to conclude that ∠APQ = ∠CQD and ∠AQP = ∠DPC.

Step 3: Now, let's consider the triangles APQ and CQD. We know that ∠APQ = ∠CQD (from Step 2) and ∠AQP = ∠DPC (from Step 2). Additionally, we have AD || BC, which implies that ∠DAQ = ∠CBQ (alternate interior angles).

Step 4: By Angle-Angle (AA) similarity, we can conclude that triangles APQ and CQD are similar.

Step 5: Since triangles APQ and CQD are similar, their corresponding sides are proportional. In particular, we have AP/CQ = AQ/QD.

Step 6: Now, let's consider the diagonals of the parallelogram ABCD. The diagonals AC and BD bisect each other at point O.

Step 7: Since the diagonals of a parallelogram bisect each other, we have AO = CO and BO = DO.

Step 8: From Step 7, we can conclude that triangles APO and CQO are congruent by Side-Angle-Side (SAS) congruence. We have AO = CO (from Step 7), ∠AOP = ∠COQ (vertical angles), and OP = OQ (common side).

Step 9: By congruence, we have AP = CQ (corresponding parts of congruent triangles).

Therefore, we have proved that AP = CQ in the parallelogram ABCD when a line intersects sides AD and BC at points P and Q respectively.