Question

In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $$DP = BQ$$ (refer to the given figure). Show that :
(i) $$\triangle APD \cong \triangle CQB$$
(ii) $$AP=CQ$$
(iii) $$\triangle AQB \cong \triangle CPD$$
(iv) $$AQ=CP$$
(v) APCQ is a parallelogram

Answer 1

In parallelogram $$ABCD$$, two points $$P$$ and $$Q$$ are taken on diagonal $$BD$$ such that

$$DP=BQ$$

proof:(i)In $$\triangle APD$$ and $$\triangle {CQB}$$,

$$\therefore AD\parallel BC$$

opposite sides of parallelogram $$ABCD$$ and a transversal $$BD$$ intersect them

$$\therefore \angle{ ABD}=\angle{ CBD}$$ alternate interior angles

$$\Rightarrow \angle ADP=\angle CBQ$$ ....$$(1)$$

$$ DP=BC$$ ....$$(2)$$

$$ AD=CB$$......$$(3)$$

opposite sides of parallelogram $$ABCD$$

From $$\left(1\right) , \left(2\right)$$ and $$\left(3\right)$$

$$\triangle APD\cong\triangle CQB$$ by SAS congruence

(ii)$$\therefore \triangle APD =\triangle CQB$$ proved in $$(1)$$ above

$$\therefore AP=CQ$$

(iii)In $$\triangle AQB$$ and $$\triangle CPD$$

$$\therefore AB||CD$$ since opposite sides of parallelogram $$ABCD$$ and a transversal $$BD$$ intersects them

$$\therefore \angle ABD=\angle CDB$$ as alternate interior angles

$$\Rightarrow \angle ABQ=\angle CDP$$

$$\Rightarrow QB=PD$$ (given)

$$\Rightarrow AB=CD$$ opposite sides of parallelogram $$ABCD$$

$$\therefore \triangle {AQB}\cong\triangle{ CPD}$$ by SAS congruence rule

$$(iv)\therefore \triangle AQB=\triangle CPD$$ proved in $$\left(iii\right)$$ above

$$\therefore AQ=CP $$ by CPCT

$$(v)\therefore$$ the diagonals of a parallelogram bisects each other

$$\therefore OB=OD$$

$$\therefore OB-BQ=OD-DP$$

$$\therefore BQ=DP$$ (given)

$$\therefore OQ=OP.....(1)$$

Also, $$OA=OC.......(2) $$

$$\therefore$$ diagonals of a parallelogram bisects each other

From $$(1)$$ and $$(2), APCQ$$ is a parallelogram