Question

ABCD is a parallelogram in which AP bisects angle A and BQ bisects angle B. Prove that: (i)AQ=BP (ii)PQ=CD (iii)ABPQ is a parallelogram

Answer 1

this is your answer


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Answer 2

ABPQ & ABCD are PARALLELOGRAM

Step-by-step explanation:

Let $\angle QAB = 2x$

$\angle ABP = 180 - 2x$

$\angle PAB = x$

$\angle APB = 180 - (x + 180 - 2x)$

= 180 - (180 - x)

= 180 - 180 + x

= x

In $\Delta BAP,$

$\angle BAP = \angle APB$

or AB = BP """"(1)

Now, $\angle PBA = 180 - 2x$

$\angle PBQ = \angle QBA$

$\angle QBA = \frac {180-2x}{2}$

= 90 - x  

So,$\angle PBQ = 90 - x$

and $\angle PBQ = \angle AQB$ (Alternate Interior Angles)

$\angle AQB = 90 - x$

$\angle QBA = \angle AQB$

hence, AB = AQ """(2)

From equations (1) & (2) we get-

(i) AQ = PB Proved !

(iii) Since, AQ = BP & $AD \parallel CB$ (As ABCD is Parallelogram)

$AQ \parallel PB$ (Opposiote sides are equal & parallel, So, ABPQ is a parallelogram)

(ii) AB = PQ (As  ABPQ is a parallelogram)

and AB = CD (As  ABCD is a parallelogram)

Hence, CD = PQ (Proved)