Question

Proofs that angle bisector of a parallelogram form a rectangle.

Answer 1

Given: ABCD is a parallelogram. AE bisects $\angle BAD$. BF bisects $\angle ABC$. CG bisects $\angle BCD$ and DH bisects $\angle ADC$
To prove: LKJI is a rectangle
$\angle BAD+\angle ABC={180}^{\circ }$ because adjacent angles of a parallelogram are supplementary
$\angle BAJ=\frac{1}{2}\angle BADsinceAEbisects\angle BAD\angle ABJ=\frac{1}{2}\angle ABCsinceDHbisects\angle ABC\angle BAJ+\angle ABJ=\frac{1}{2}\angle BAD+\frac{1}{2}\angle ABC$
$=\frac{1}{2}[\angle BAD+\angle ABC]=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }$
[Since sum of adjacent angles of a parallelogram are suppliementary]
$\Delta ABJ$ is a right triangle since its acute interior angles are complementary
Similar in $\Delta CDL$ we get $\angle DLC={90}^{\circ }$ and in $\Delta ADI$ we get $\angle AID={90}^{\circ }$
Then $\angle JIL={90}^{\circ }$ as $\angle AID$ and $\angle JIL$ are vertical opposite angles of quadrilateral LKJI are right angles, hence 4th angle is also a right angle.
Thus LKJI is a rectangle