Question

ABCD is a rectangle in which DP and BQ are perpendiculars from D and B respectively on diagonal AC.

Show that
(i)$\Delta ADP \cong \Delta CBQ$
(ii)$\angle ADP = \angle CBQ$
(iii)DP=BQ

Answer 1

Hint: ABCD is rectangle which means its opposite sides are equal. We can prove $\Delta ADP \cong \Delta CBQ$ by using AAS criterion according to which if two angles and one side of a triangle are equal to the two angles and one side of another triangle then the triangles are congruent. Then by CPCT we can prove $\angle ADP = \angle CBQ$ and side DP=BQ .

Complete step-by-step answer:
Given ABCD is a rectangle then AB=CD and AD=BC
And DP and BQ are perpendiculars from D and B respectively on diagonal AC.
Then $\angle P = \angle Q = {90^ \circ }$ and two triangles $\Delta ADP$ and $\Delta CBQ$ are formed.

Now we have to prove-
(i)We have to prove that $\Delta ADP$ and $\Delta CBQ$are congruent.
In $\Delta ADP$ and $\Delta CBQ$ ,
$\angle APD = \angle CQB$$ = {90^ \circ }$ (Given)
$\angle DAP = \angle BCQ$ (Because they are alternate angles)
Since opposite sides of rectangle are equal so we can write,
Side AB=side BC
Hence by Angle-Angle-Side congruence
$\Delta ADP \cong \Delta CBQ$ Hence Proved
(ii) We have to prove that$\angle ADP = \angle CBQ$
Since we have already proved that$\Delta ADP \cong \Delta CBQ$
We know that corresponding parts of two congruent triangles are always equal.
Then by CPCT (Corresponding Parts of Congruent Triangles)
$\angle ADP = \angle CBQ$ Hence Proved
(iii)We have to prove that side DP=side BQ
Since we have already proved that $\Delta ADP \cong \Delta CBQ$
We know that corresponding parts of two congruent triangles are always equal.
Then by CPCT (Corresponding Parts of Congruent Triangles)
Side DP=side BQ Hence Proved.

Note: AAS is one of the triangle congruence criterion which is used to prove that two triangles have equal sides and equal angles. The other congruence criterion are- SSS(side-side-side), ASA(angle-side-angle), SAS(side-angle-side) and HL(hypotenuse-leg).We cannot use ASA criterion because AD and BC are not include between the equal angles.