Q20 of 109 Page 330

In the adjoining figure, ABCD and PQRC are rectangles, where Q is the mid-point of AC.

Prove that (i) DP = PC (ii)


(i) Here, we have

∠CRQ = ∠CBA = 90o


Thus, RQ ∣∣​ AB


Now, In ∆ABC,



Q is the mid-point of AC and QR ∣∣​ AB​.


Thus, R is the mid-point of BC.


In the same way, P is the midpoint of DC.


Hence, DP = PC


(ii) Here, let us join B to D.


Now, In ∆​CDB,


P and R are the mid points of DC and BC respectively.


Since, AC = BD


Thus, PR ∣∣​ DB and PR = DB = AC


More from this chapter

All 109 →
16

Show that a diagonal divides a parallelogram into two triangles of equal area.

 17

In the given figure, AC is a diagonal of quad. ABCD in which BL ⊥ AC and DM ⊥ AC. Prove that or (quad. ABCD)

 18

||gm ABCD and rectangle ABEF have the same base AB and are equal in areas. Show that the perimeter of the ||gm is greater than that of the rectangle.

 19

In the adjoining figure, ABCD is a ||gm and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF = 2AB.