Question

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A transversal cuts two parallel lines at A and B. The two interior angles at A are bisected and so are the two interior angles at B; the four bisectors form a quadrilateral ACBD.

•Prove that :-

() ACBD is a rectangle.
(1) CD is parallel to the original parallel lines.


@MrMagician
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Left~​

Answer 1

Answer:

Explanation:

Proof: RS, PS, PQ and RQ are bisectors of interior angles formed by the transversal with the parallel lines. Therefore quadrilateral PQRS is a parallelogram as both the pairs of opposite sides are parallel. That is PQRS is a parallelogram and one of the angle is a right angle.

Answer 2

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$\text { (1) } \angle \mathrm{LAB}+\angle \mathrm{BAM}=180^{\circ} \quad[\text { LAM is a straight line }]$

$\begin{aligned} &1 / 2(\angle \mathrm{LAB}+\angle \mathrm{BAM})=90^{\circ}\\ &1 / 2 \angle \mathrm{LAB}+1 / 2 \angle \mathrm{BAM}=90^{\circ}\\ &\angle 2+\angle 3=90^{\circ}\\ &\angle C A D=90^{\circ}\\ &\angle \mathrm{A}=90^{\circ} \end{aligned}$

$\begin{aligned} &\text { (2) Similarly, } \angle \mathrm{PBA}+\angle \mathrm{QBA}=180^{\circ} \quad[\mathrm{PBQ} \text { is a straight line] }\\ &1 / 2(\angle \mathrm{PBA}+\angle \mathrm{QBA})=90^{\circ}\\ &1 / 2 \angle P B A+1 / 2 \angle Q B A=90^{\circ}\\ &\angle 6+\angle 7=90^{\circ}\\ &\angle \mathrm{CBD}=90^{\circ}\\ &\angle \mathrm{B}=90^{\circ} \end{aligned}$

$\begin{aligned} &\text { (3) } \angle \mathrm{LAB}+\angle \mathrm{ABP}=180^{\circ} \quad \text { [Sum of co-interior angles is } 180^{\circ} \text { and given } \left.\mathrm{LM} \| \mathrm{PQ}\right]\\ &1 / 2 \angle L A B+1 / 2 \angle A B P=90^{\circ}\\ &\angle 2+\angle 6=90^{\circ} \quad[\text { Since, } \mathrm{AC} \text { and } \mathrm{BC} \text { is bisector of } \angle \mathrm{LAB} \& \angle \mathrm{PBA} \text { respectively }] \end{aligned}$

❝Hope It Helps❞