Q11 of 95 Page 15

In Fig. 14.98, AB=AC and CP||BA and AP is the bisector of exterior ∠CAD of ΔABC. Prove that (i) ∠PAC=∠BCA (ii) ABCP is a parallelogram.

Given,


In figure 14.98


AB = AC


CP ⎸⎸BA


AP is bisector of exterior angle ∠CAD


AB = AC


∠C = ∠B


NOW,


∠CAD = ∠B +∠C


2∠CAP = 2∠C


∠CAP = ∠C


AP ⎸⎸BC


But, AB ⎸⎸CP (given)


Hence ABCP is a parallelogram


More from this chapter

All 95 →
9

ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ΔPQR is double the perimeter of ΔABC.

 10

In Fig. 14.97, BE⊥AC. AD is any line from A to BC intersecting BE in H, P, Q and R are respectively the mid points of AH, AB and BC. Prove that ∠PQR = 90°

 12

ABCD is a kite having AB=AD and BC=CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.

 13

Let ABC be an isosceles triangle in which AB=AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.