Question

4 in parallelogram abcd on the right, draw one straight line passing through intersection point o of the diagonal lines, and let the intersection points of sides ab and dc be p and q respectively. in this case, prove that ap = cq as follows. fill the appropriate words in the blanks. proof in △aop and △, since the property of a parallelogram, = …………① ∠pao = ∠ …………② (the alternate angles are equal ) ∠poa = ∠ …………③ (the opposite angles are equal ) from ①, ②, and ③, since the two and the included are equal respectively, △ ≅ △. therefore, =.

Explanation:

Step1: Identify the triangles

In $\triangle AOP$ and $\triangle COQ$.

Step2: Use parallelogram property

Since in parallelogram \(ABCD\), \(AO = CO\) (diagonals bisect each other).

Step3: Alternate angles

\(\angle PAO=\angle QCO\) (alternate angles as \(AB\parallel DC\)).

Step4: Opposite angles

\(\angle POA=\angle QOC\) (opposite angles).

Step5: ASA congruence

From above, since two angles and the included side are equal respectively, \(\triangle AOP\cong\triangle COQ\) (by ASA - Angle - Side - Angle congruence criterion).

Step6: Corresponding parts

Therefore, \(AP = CQ\) (corresponding parts of congruent triangles).

Answer:

In \(\triangle AOP\) and \(\triangle COQ\), \(AO = CO\), \(\angle PAO=\angle QCO\), \(\angle POA=\angle QOC\), two angles and the included side, \(\triangle AOP\cong\triangle COQ\), \(AP = CQ\)