Question

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

Answer 1

Step-by-step explanation:

as I explained may it helps you and plzz.. follow me.

Answer 2

Answer:

Step-by-step explanation:

part (1)

EXPLANATION:

Given :

LM ||PQ and AB is the transversal line cuttting ∠M at A and PQ at B

AC,AD, BC and BD  is the bisector ∠LAB , ∠BAM , ∠PAB and∠ABQ respectively.

AC  and BC  intersect at C

AD  and BD  intersect at D

A quadrilateral ABCD is formed

Step 1 of 6

∠LAB +∠BAM = 180 (linear pair)

$\frac{1}{2}$(∠LAB +∠BAM) = 90

$\frac{1}{2}$∠LAB + $\frac{1}{2}$∠BAM = 90

∠2 +∠3 = 90

but given that AC and AD is bisector of ∠LAB and ∠BAM

So , ∠CAD = 90

and ∠A = 90

Step 2 of6

Similarly , ∠PBA +∠QBA = 180

$\frac{1}{2}$∠ PBA +$\frac{1}{2}$∠QBA = 90

∠ 6 +∠7 = 90

but BC and BD is bisector of ∠ PBA and ∠QBA

∠CBD = 90

∠B = 90

Step 3 of 6

∠LAB +∠ABP = 180       [Co - interior angle ]

from given LM || PQ

$\frac{1}{2}$∠LAB +$\frac{1}{2}$∠ABP = 90

∠2 +∠6 = 90

but AC and BC is bisector of ∠LAB and ∠PBA

In Δ ACB ,

∠2 +∠6 +∠C = 180

(∠2 +∠6) +∠C = 180

90 + ∠C = 180

∠C = 180- 90 =90

∠C = 90

Step 4 of6

∠MAB +∠ABQ = 180

and given LM||PQ    $\frac{1}{2}$∠MAB+$\frac{1}{2}$∠ABQ=90

∠3+∠7 = 90

but AD and BD is bisector of ∠MAB and ∠ABQ

In ΔADB ,

∠3 +∠7 +∠D = 180

(∠3+∠7) +∠D = 180

90 +∠D = 180

∠D = 180 - 90

∠D = 90

Step 5 of6

∠LAB +∠BAM = 180

∠BAM =∠ABP       (From step 1 and 2)

$\frac{1}{2}$∠BAM =$\frac{1}{2}$∠ABP

∠3+∠6

but , AD and BC is bisector of ∠BAM and ∠ABP

Similarly ,∠2 =∠7

In ΔABC and ΔABD ,

∠2=∠7

AB = AB      (Common)

∠6 =∠3 (from 7)

By ASA property of congurency

ΔABC≅ΔABD

thus by CPCT we get

AC=DB

Also , CB = AD

Step 6 of6

∠A =∠B=∠C=∠D = 90 °

from step 1,2,3 and 4

AC = DB (from step 5)

CB = AD (from step 5)

Final answer:

Hence ,ABCD is a rectangle.

part (ii)

Explanation :

Given :

LM ||PQ and AB is the transversal line cuttting ∠M at A and PQ at B

AC,AD, BC and BD  is the bisector ∠LAB , ∠BAM , ∠PAB and∠ABQ respectively.

AC  and BC  intersect at C

AD  and BD  intersect at D

A quadrilateral ABCD is formed

Step 1 of 2:

ABCD is a rectangle we proved in part (i)

OA = OD          (diagonal of rectangle )

In ΔAOD , we have

OA = OD

∠9 =∠3  (angle opposite to equal sides)   ........(i)

∠3=∠4 (AD bisects ∠MAb )  .................(ii)

∠9 =∠4  ................(from i and ii)

but these are alternate angles .

OD|| LM

⇒CD || LM

Step 2 of 2

Similarly , we can prove that

∠10 =∠8

but these are alternate angles ,

so , OD || PQ

⇒CD || LM .................... from (iii)

CD|| PQ

Final answer :

Hence we prove that CD is parallel to the original parallel lines.